I am looking to add a simple search field, would like to use something like
collectionRef.where('name', 'contains', 'searchTerm')
I tried using where('name', '==', '%searchTerm%'), but it didn't return anything.
I agree with #Kuba's answer, But still, it needs to add a small change to work perfectly for search by prefix. here what worked for me
For searching records starting with name queryText
collectionRef
.where('name', '>=', queryText)
.where('name', '<=', queryText+ '\uf8ff')
The character \uf8ff used in the query is a very high code point in the Unicode range (it is a Private Usage Area [PUA] code). Because it is after most regular characters in Unicode, the query matches all values that start with queryText.
Full-Text Search, Relevant Search, and Trigram Search!
UPDATE - 2/17/21 - I created several new Full Text Search Options.
See Code.Build for details.
Also, side note, dgraph now has websockets for realtime... wow, never saw that coming, what a treat! Cloud Dgraph - Amazing!
--Original Post--
A few notes here:
1.) \uf8ff works the same way as ~
2.) You can use a where clause or start end clauses:
ref.orderBy('title').startAt(term).endAt(term + '~');
is exactly the same as
ref.where('title', '>=', term).where('title', '<=', term + '~');
3.) No, it does not work if you reverse startAt() and endAt() in every combination, however, you can achieve the same result by creating a second search field that is reversed, and combining the results.
Example: First you have to save a reversed version of the field when the field is created. Something like this:
// collection
const postRef = db.collection('posts')
async function searchTitle(term) {
// reverse term
const termR = term.split("").reverse().join("");
// define queries
const titles = postRef.orderBy('title').startAt(term).endAt(term + '~').get();
const titlesR = postRef.orderBy('titleRev').startAt(termR).endAt(termR + '~').get();
// get queries
const [titleSnap, titlesRSnap] = await Promise.all([
titles,
titlesR
]);
return (titleSnap.docs).concat(titlesRSnap.docs);
}
With this, you can search the last letters of a string field and the first, just not random middle letters or groups of letters. This is closer to the desired result. However, this won't really help us when we want random middle letters or words. Also, remember to save everything lowercase, or a lowercase copy for searching, so case won't be an issue.
4.) If you have only a few words, Ken Tan's Method will do everything you want, or at least after you modify it slightly. However, with only a paragraph of text, you will exponentially create more than 1MB of data, which is bigger than firestore's document size limit (I know, I tested it).
5.) If you could combine array-contains (or some form of arrays) with the \uf8ff trick, you might could have a viable search that does not reach the limits. I tried every combination, even with maps, and a no go. Anyone figures this out, post it here.
6.) If you must get away from ALGOLIA and ELASTIC SEARCH, and I don't blame you at all, you could always use mySQL, postSQL, or neo4Js on Google Cloud. They are all 3 easy to set up, and they have free tiers. You would have one cloud function to save the data onCreate() and another onCall() function to search the data. Simple...ish. Why not just switch to mySQL then? The real-time data of course! When someone writes DGraph with websocks for real-time data, count me in!
Algolia and ElasticSearch were built to be search-only dbs, so there is nothing as quick... but you pay for it. Google, why do you lead us away from Google, and don't you follow MongoDB noSQL and allow searches?
There's no such operator, allowed ones are ==, <, <=, >, >=.
You can filter by prefixes only, for example for everything that starts between bar and foo you can use
collectionRef
.where('name', '>=', 'bar')
.where('name', '<=', 'foo')
You can use external service like Algolia or ElasticSearch for that.
While Kuba's answer is true as far as restrictions go, you can partially emulate this with a set-like structure:
{
'terms': {
'reebok': true,
'mens': true,
'tennis': true,
'racket': true
}
}
Now you can query with
collectionRef.where('terms.tennis', '==', true)
This works because Firestore will automatically create an index for every field. Unfortunately this doesn't work directly for compound queries because Firestore doesn't automatically create composite indexes.
You can still work around this by storing combinations of words but this gets ugly fast.
You're still probably better off with an outboard full text search.
While Firebase does not explicitly support searching for a term within a string,
Firebase does (now) support the following which will solve for your case and many others:
As of August 2018 they support array-contains query. See: https://firebase.googleblog.com/2018/08/better-arrays-in-cloud-firestore.html
You can now set all of your key terms into an array as a field then query for all documents that have an array that contains 'X'. You can use logical AND to make further comparisons for additional queries. (This is because firebase does not currently natively support compound queries for multiple array-contains queries so 'AND' sorting queries will have to be done on client end)
Using arrays in this style will allow them to be optimized for concurrent writes which is nice! Haven't tested that it supports batch requests (docs don't say) but I'd wager it does since its an official solution.
Usage:
collection("collectionPath").
where("searchTermsArray", "array-contains", "term").get()
Per the Firestore docs, Cloud Firestore doesn't support native indexing or search for text fields in documents. Additionally, downloading an entire collection to search for fields client-side isn't practical.
Third-party search solutions like Algolia and Elastic Search are recommended.
I'm sure Firebase will come out with "string-contains" soon to capture any index[i] startAt in the string...
But
I’ve researched the webs and found this solution thought of by someone else
set up your data like this
state = { title: "Knitting" };
// ...
const c = this.state.title.toLowerCase();
var array = [];
for (let i = 1; i < c.length + 1; i++) {
array.push(c.substring(0, i));
}
firebase
.firestore()
.collection("clubs")
.doc(documentId)
.update({
title: this.state.title,
titleAsArray: array
});
query like this
firebase.firestore()
.collection("clubs")
.where(
"titleAsArray",
"array-contains",
this.state.userQuery.toLowerCase()
)
As of today (18-Aug-2020), there are basically 3 different workarounds, which were suggested by the experts, as answers to the question.
I have tried them all. I thought it might be useful to document my experience with each one of them.
Method-A: Using: (dbField ">=" searchString) & (dbField "<=" searchString + "\uf8ff")
Suggested by #Kuba & #Ankit Prajapati
.where("dbField1", ">=", searchString)
.where("dbField1", "<=", searchString + "\uf8ff");
A.1 Firestore queries can only perform range filters (>, <, >=, <=) on a single field. Queries with range filters on multiple fields are not supported. By using this method, you can't have a range operator in any other field on the db, e.g. a date field.
A.2. This method does NOT work for searching in multiple fields at the same time. For example, you can't check if a search string is in any of the fileds (name, notes & address).
Method-B: Using a MAP of search strings with "true" for each entry in the map, & using the "==" operator in the queries
Suggested by #Gil Gilbert
document1 = {
'searchKeywordsMap': {
'Jam': true,
'Butter': true,
'Muhamed': true,
'Green District': true,
'Muhamed, Green District': true,
}
}
.where(`searchKeywordsMap.${searchString}`, "==", true);
B.1 Obviously, this method requires extra processing every time data is saved to the db, and more importantly, requires extra space to store the map of search strings.
B.2 If a Firestore query has a single condition like the one above, no index needs to be created beforehand. This solution would work just fine in this case.
B.3 However, if the query has another condition, e.g. (status === "active",) it seems that an index is required for each "search string" the user enters. In other words, if a user searches for "Jam" and another user searches for "Butter", an index should be created beforehand for the string "Jam", and another one for "Butter", etc. Unless you can predict all possible users' search strings, this does NOT work - in case of the query has other conditions!
.where(searchKeywordsMap["Jam"], "==", true); // requires an index on searchKeywordsMap["Jam"]
.where("status", "==", "active");
**Method-C: Using an ARRAY of search strings, & the "array-contains" operator
Suggested by #Albert Renshaw & demonstrated by #Nick Carducci
document1 = {
'searchKeywordsArray': [
'Jam',
'Butter',
'Muhamed',
'Green District',
'Muhamed, Green District',
]
}
.where("searchKeywordsArray", "array-contains", searchString);
C.1 Similar to Method-B, this method requires extra processing every time data is saved to the db, and more importantly, requires extra space to store the array of search strings.
C.2 Firestore queries can include at most one "array-contains" or "array-contains-any" clause in a compound query.
General Limitations:
None of these solutions seems to support searching for partial strings. For example, if a db field contains "1 Peter St, Green District", you can't search for the string "strict."
It is almost impossible to cover all possible combinations of expected search strings. For example, if a db field contains "1 Mohamed St, Green District", you may NOT be able to search for the string "Green Mohamed", which is a string having the words in a different order than the order used in the DB field.
There is no one solution that fits all. Each workaround has its limitations. I hope the information above can help you during the selection process between these workarounds.
For a list of Firestore query conditions, please check out the documentation https://firebase.google.com/docs/firestore/query-data/queries.
I have not tried https://fireblog.io/blog/post/firestore-full-text-search, which is suggested by #Jonathan.
Late answer but for anyone who's still looking for an answer, Let's say we have a collection of users and in each document of the collection we have a "username" field, so if want to find a document where the username starts with "al" we can do something like
FirebaseFirestore.getInstance()
.collection("users")
.whereGreaterThanOrEqualTo("username", "al")
I used trigram just like Jonathan said it.
trigrams are groups of 3 letters stored in a database to help with searching. so if I have data of users and I let' say I want to query 'trum' for donald trump I have to store it this way
and I just to recall this way
onPressed: () {
//LET SAY YOU TYPE FOR 'tru' for trump
List<String> search = ['tru', 'rum'];
Future<QuerySnapshot> inst = FirebaseFirestore.instance
.collection("users")
.where('trigram', arrayContainsAny: search)
.get();
print('result=');
inst.then((value) {
for (var i in value.docs) {
print(i.data()['name']);
}
});
that will get correct result no matter what
EDIT 05/2021:
Google Firebase now has an extension to implement Search with Algolia. Algolia is a full text search platform that has an extensive list of features. You are required to have a "Blaze" plan on Firebase and there are fees associated with Algolia queries, but this would be my recommended approach for production applications. If you prefer a free basic search, see my original answer below.
https://firebase.google.com/products/extensions/firestore-algolia-search
https://www.algolia.com
ORIGINAL ANSWER:
The selected answer only works for exact searches and is not natural user search behavior (searching for "apple" in "Joe ate an apple today" would not work).
I think Dan Fein's answer above should be ranked higher. If the String data you're searching through is short, you can save all substrings of the string in an array in your Document and then search through the array with Firebase's array_contains query. Firebase Documents are limited to 1 MiB (1,048,576 bytes) (Firebase Quotas and Limits) , which is about 1 million characters saved in a document (I think 1 character ~= 1 byte). Storing the substrings is fine as long as your document isn't close to 1 million mark.
Example to search user names:
Step 1: Add the following String extension to your project. This lets you easily break up a string into substrings. (I found this here).
extension String {
var length: Int {
return count
}
subscript (i: Int) -> String {
return self[i ..< i + 1]
}
func substring(fromIndex: Int) -> String {
return self[min(fromIndex, length) ..< length]
}
func substring(toIndex: Int) -> String {
return self[0 ..< max(0, toIndex)]
}
subscript (r: Range<Int>) -> String {
let range = Range(uncheckedBounds: (lower: max(0, min(length, r.lowerBound)),
upper: min(length, max(0, r.upperBound))))
let start = index(startIndex, offsetBy: range.lowerBound)
let end = index(start, offsetBy: range.upperBound - range.lowerBound)
return String(self[start ..< end])
}
Step 2: When you store a user's name, also store the result of this function as an array in the same Document. This creates all variations of the original text and stores them in an array. For example, the text input "Apple" would creates the following array: ["a", "p", "p", "l", "e", "ap", "pp", "pl", "le", "app", "ppl", "ple", "appl", "pple", "apple"], which should encompass all search criteria a user might enter. You can leave maximumStringSize as nil if you want all results, however, if there is long text, I would recommend capping it before the document size gets too big - somewhere around 15 works fine for me (most people don't search long phrases anyway).
func createSubstringArray(forText text: String, maximumStringSize: Int?) -> [String] {
var substringArray = [String]()
var characterCounter = 1
let textLowercased = text.lowercased()
let characterCount = text.count
for _ in 0...characterCount {
for x in 0...characterCount {
let lastCharacter = x + characterCounter
if lastCharacter <= characterCount {
let substring = textLowercased[x..<lastCharacter]
substringArray.append(substring)
}
}
characterCounter += 1
if let max = maximumStringSize, characterCounter > max {
break
}
}
print(substringArray)
return substringArray
}
Step 3: You can use Firebase's array_contains function!
[yourDatabasePath].whereField([savedSubstringArray], arrayContains: searchText).getDocuments....
I just had this problem and came up with a pretty simple solution.
String search = "ca";
Firestore.instance.collection("categories").orderBy("name").where("name",isGreaterThanOrEqualTo: search).where("name",isLessThanOrEqualTo: search+"z")
The isGreaterThanOrEqualTo lets us filter out the beginning of our search and by adding a "z" to the end of the isLessThanOrEqualTo we cap our search to not roll over to the next documents.
I actually think the best solution to do this within Firestore is to put all substrings in an array, and just do an array_contains query. This allows you to do substring matching. A bit overkill to store all substrings but if your search terms are short it's very very reasonable.
If you don't want to use a third-party service like Algolia, Firebase Cloud Functions are a great alternative. You can create a function that can receive an input parameter, process through the records server-side and then return the ones that match your criteria.
This worked for me perfectly but might cause performance issues.
Do this when querying firestore:
Future<QuerySnapshot> searchResults = collectionRef
.where('property', isGreaterThanOrEqualTo: searchQuery.toUpperCase())
.getDocuments();
Do this in your FutureBuilder:
return FutureBuilder(
future: searchResults,
builder: (context, snapshot) {
List<Model> searchResults = [];
snapshot.data.documents.forEach((doc) {
Model model = Model.fromDocumet(doc);
if (searchQuery.isNotEmpty &&
!model.property.toLowerCase().contains(searchQuery.toLowerCase())) {
return;
}
searchResults.add(model);
})
};
Following code snippet takes input from user and acquires data starting with the typed one.
Sample Data:
Under Firebase Collection 'Users'
user1: {name: 'Ali', age: 28},
user2: {name: 'Khan', age: 30},
user3: {name: 'Hassan', age: 26},
user4: {name: 'Adil', age: 32}
TextInput: A
Result:
{name: 'Ali', age: 28},
{name: 'Adil', age: 32}
let timer;
// method called onChangeText from TextInput
const textInputSearch = (text) => {
const inputStart = text.trim();
let lastLetterCode = inputStart.charCodeAt(inputStart.length-1);
lastLetterCode++;
const newLastLetter = String.fromCharCode(lastLetterCode);
const inputEnd = inputStart.slice(0,inputStart.length-1) + lastLetterCode;
clearTimeout(timer);
timer = setTimeout(() => {
firestore().collection('Users')
.where('name', '>=', inputStart)
.where('name', '<', inputEnd)
.limit(10)
.get()
.then(querySnapshot => {
const users = [];
querySnapshot.forEach(doc => {
users.push(doc.data());
})
setUsers(users); // Setting Respective State
});
}, 1000);
};
2021 Update
Took a few things from other answers. This one includes:
Multi word search using split (acts as OR)
Multi key search using flat
A bit limited on case-sensitivity, you can solve this by storing duplicate properties in uppercase. Ex: query.toUpperCase() user.last_name_upper
// query: searchable terms as string
let users = await searchResults("Bob Dylan", 'users');
async function searchResults(query = null, collection = 'users', keys = ['last_name', 'first_name', 'email']) {
let querySnapshot = { docs : [] };
try {
if (query) {
let search = async (query)=> {
let queryWords = query.trim().split(' ');
return queryWords.map((queryWord) => keys.map(async (key) =>
await firebase
.firestore()
.collection(collection)
.where(key, '>=', queryWord)
.where(key, '<=', queryWord + '\uf8ff')
.get())).flat();
}
let results = await search(query);
await (await Promise.all(results)).forEach((search) => {
querySnapshot.docs = querySnapshot.docs.concat(search.docs);
});
} else {
// No query
querySnapshot = await firebase
.firestore()
.collection(collection)
// Pagination (optional)
// .orderBy(sortField, sortOrder)
// .startAfter(startAfter)
// .limit(perPage)
.get();
}
} catch(err) {
console.log(err)
}
// Appends id and creates clean Array
const items = [];
querySnapshot.docs.forEach(doc => {
let item = doc.data();
item.id = doc.id;
items.push(item);
});
// Filters duplicates
return items.filter((v, i, a) => a.findIndex(t => (t.id === v.id)) === i);
}
Note: the number of Firebase calls is equivalent to the number of words in the query string * the number of keys you're searching on.
Same as #nicksarno but with a more polished code that doesn't need any extension:
Step 1
func getSubstrings(from string: String, maximumSubstringLenght: Int = .max) -> [Substring] {
let string = string.lowercased()
let stringLength = string.count
let stringStartIndex = string.startIndex
var substrings: [Substring] = []
for lowBound in 0..<stringLength {
for upBound in lowBound..<min(stringLength, lowBound+maximumSubstringLenght) {
let lowIndex = string.index(stringStartIndex, offsetBy: lowBound)
let upIndex = string.index(stringStartIndex, offsetBy: upBound)
substrings.append(string[lowIndex...upIndex])
}
}
return substrings
}
Step 2
let name = "Lorenzo"
ref.setData(["name": name, "nameSubstrings": getSubstrings(from: name)])
Step 3
Firestore.firestore().collection("Users")
.whereField("nameSubstrings", arrayContains: searchText)
.getDocuments...
With Firestore you can implement a full text search but it will still cost more reads than it would have otherwise, and also you'll need to enter and index the data in a particular way, So in this approach you can use firebase cloud functions to tokenise and then hash your input text while choosing a linear hash function h(x) that satisfies the following - if x < y < z then h(x) < h (y) < h(z). For tokenisation you can choose some lightweight NLP Libraries in order to keep the cold start time of your function low that can strip unnecessary words from your sentence. Then you can run a query with less than and greater than operator in Firestore.
While storing your data also, you'll have to make sure that you hash the text before storing it, and store the plain text also as if you change the plain text the hashed value will also change.
Typesense service provide substring search for Firebase Cloud Firestore database.
https://typesense.org/docs/guide/firebase-full-text-search.html
Following is the relevant codes of typesense integration for my project.
lib/utils/typesense.dart
import 'dart:convert';
import 'package:flutter_instagram_clone/model/PostModel.dart';
import 'package:http/http.dart' as http;
class Typesense {
static String baseUrl = 'http://typesense_server_ip:port/';
static String apiKey = 'xxxxxxxx'; // your Typesense API key
static String resource = 'collections/postData/documents/search';
static Future<List<PostModel>> search(String searchKey, int page, {int contentType=-1}) async {
if (searchKey.isEmpty) return [];
List<PostModel> _results = [];
var header = {'X-TYPESENSE-API-KEY': apiKey};
String strSearchKey4Url = searchKey.replaceFirst('#', '%23').replaceAll(' ', '%20');
String url = baseUrl +
resource +
'?q=${strSearchKey4Url}&query_by=postText&page=$page&sort_by=millisecondsTimestamp:desc&num_typos=0';
if(contentType==0)
{
url += "&filter_by=isSelling:false";
} else if(contentType == 1)
{
url += "&filter_by=isSelling:true";
}
var response = await http.get(Uri.parse(url), headers: header);
var data = json.decode(response.body);
for (var item in data['hits']) {
PostModel _post = PostModel.fromTypeSenseJson(item['document']);
if (searchKey.contains('#')) {
if (_post.postText.toLowerCase().contains(searchKey.toLowerCase()))
_results.add(_post);
} else {
_results.add(_post);
}
}
print(_results.length);
return _results;
}
static Future<List<PostModel>> getHubPosts(String searchKey, int page,
{List<String>? authors, bool? isSelling}) async {
List<PostModel> _results = [];
var header = {'X-TYPESENSE-API-KEY': apiKey};
String filter = "";
if (authors != null || isSelling != null) {
filter += "&filter_by=";
if (isSelling != null) {
filter += "isSelling:$isSelling";
if (authors != null && authors.isNotEmpty) {
filter += "&&";
}
}
if (authors != null && authors.isNotEmpty) {
filter += "authorID:$authors";
}
}
String url = baseUrl +
resource +
'?q=${searchKey.replaceFirst('#', '%23')}&query_by=postText&page=$page&sort_by=millisecondsTimestamp:desc&num_typos=0$filter';
var response = await http.get(Uri.parse(url), headers: header);
var data = json.decode(response.body);
for (var item in data['hits']) {
PostModel _post = PostModel.fromTypeSenseJson(item['document']);
_results.add(_post);
}
print(_results.length);
return _results;
}
}
lib/services/hubDetailsService.dart
import 'package:flutter/material.dart';
import 'package:flutter_instagram_clone/model/PostModel.dart';
import 'package:flutter_instagram_clone/utils/typesense.dart';
class HubDetailsService with ChangeNotifier {
String searchKey = '';
List<String>? authors;
bool? isSelling;
int nContentType=-1;
bool isLoading = false;
List<PostModel> hubResults = [];
int _page = 1;
bool isMore = true;
bool noResult = false;
Future initSearch() async {
isLoading = true;
isMore = true;
noResult = false;
hubResults = [];
_page = 1;
List<PostModel> _results = await Typesense.search(searchKey, _page, contentType: nContentType);
for(var item in _results) {
hubResults.add(item);
}
isLoading = false;
if(_results.length < 10) isMore = false;
if(_results.isEmpty) noResult = true;
notifyListeners();
}
Future nextPage() async {
if(!isMore) return;
_page++;
List<PostModel> _results = await Typesense.search(searchKey, _page);
hubResults.addAll(_results);
if(_results.isEmpty) {
isMore = false;
}
notifyListeners();
}
Future refreshPage() async {
isLoading = true;
notifyListeners();
await initSearch();
isLoading = false;
notifyListeners();
}
Future search(String _searchKey) async {
isLoading = true;
notifyListeners();
searchKey = _searchKey;
await initSearch();
isLoading = false;
notifyListeners();
}
}
lib/ui/hub/hubDetailsScreen.dart
import 'package:flutter/cupertino.dart';
import 'package:flutter/material.dart';
import 'package:flutter_instagram_clone/constants.dart';
import 'package:flutter_instagram_clone/main.dart';
import 'package:flutter_instagram_clone/model/MessageData.dart';
import 'package:flutter_instagram_clone/model/SocialReactionModel.dart';
import 'package:flutter_instagram_clone/model/User.dart';
import 'package:flutter_instagram_clone/model/hubModel.dart';
import 'package:flutter_instagram_clone/services/FirebaseHelper.dart';
import 'package:flutter_instagram_clone/services/HubService.dart';
import 'package:flutter_instagram_clone/services/helper.dart';
import 'package:flutter_instagram_clone/services/hubDetailsService.dart';
import 'package:flutter_instagram_clone/ui/fullScreenImageViewer/FullScreenImageViewer.dart';
import 'package:flutter_instagram_clone/ui/home/HomeScreen.dart';
import 'package:flutter_instagram_clone/ui/hub/editHubScreen.dart';
import 'package:provider/provider.dart';
import 'package:smooth_page_indicator/smooth_page_indicator.dart';
class HubDetailsScreen extends StatefulWidget {
final HubModel hub;
HubDetailsScreen(this.hub);
#override
_HubDetailsScreenState createState() => _HubDetailsScreenState();
}
class _HubDetailsScreenState extends State<HubDetailsScreen> {
late HubDetailsService _service;
List<SocialReactionModel?> _reactionsList = [];
final fireStoreUtils = FireStoreUtils();
late Future<List<SocialReactionModel>> _myReactions;
final scrollController = ScrollController();
bool _isSubLoading = false;
#override
void initState() {
// TODO: implement initState
super.initState();
_service = Provider.of<HubDetailsService>(context, listen: false);
print(_service.isLoading);
init();
}
init() async {
_service.searchKey = "";
if(widget.hub.contentWords.length>0)
{
for(var item in widget.hub.contentWords) {
_service.searchKey += item + " ";
}
}
switch(widget.hub.contentType) {
case 'All':
break;
case 'Marketplace':
_service.isSelling = true;
_service.nContentType = 1;
break;
case 'Post Only':
_service.isSelling = false;
_service.nContentType = 0;
break;
case 'Keywords':
break;
}
for(var item in widget.hub.exceptWords) {
if(item == 'Marketplace') {
_service.isSelling = _service.isSelling != null?true:false;
} else {
_service.searchKey += "-" + item + "";
}
}
if(widget.hub.fromUserType == 'Followers') {
List<User> _followers = await fireStoreUtils.getFollowers(MyAppState.currentUser!.userID);
_service.authors = [];
for(var item in _followers)
_service.authors!.add(item.userID);
}
if(widget.hub.fromUserType == 'Selected') {
_service.authors = widget.hub.fromUserIds;
}
_service.initSearch();
_myReactions = fireStoreUtils.getMyReactions()
..then((value) {
_reactionsList.addAll(value);
});
scrollController.addListener(pagination);
}
void pagination(){
if(scrollController.position.pixels ==
scrollController.position.maxScrollExtent) {
_service.nextPage();
}
}
#override
Widget build(BuildContext context) {
Provider.of<HubDetailsService>(context);
PageController _controller = PageController(
initialPage: 0,
);
return Scaffold(
backgroundColor: Colors.white,
body: RefreshIndicator(
onRefresh: () async {
_service.refreshPage();
},
child: CustomScrollView(
controller: scrollController,
slivers: [
SliverAppBar(
centerTitle: false,
expandedHeight: MediaQuery.of(context).size.height * 0.25,
pinned: true,
backgroundColor: Colors.white,
title: Row(
mainAxisAlignment: MainAxisAlignment.spaceBetween,
children: [
InkWell(
onTap: (){
Navigator.pop(context);
},
child: Container(
width: 35, height: 35,
decoration: BoxDecoration(
color: Colors.white,
borderRadius: BorderRadius.circular(20)
),
child: Center(
child: Icon(Icons.arrow_back),
),
),
),
if(widget.hub.user.userID == MyAppState.currentUser!.userID)
InkWell(
onTap: () async {
var _hub = await push(context, EditHubScreen(widget.hub));
if(_hub != null) {
Navigator.pop(context, true);
}
},
child: Container(
width: 35, height: 35,
decoration: BoxDecoration(
color: Colors.white,
borderRadius: BorderRadius.circular(20)
),
child: Center(
child: Icon(Icons.edit, color: Colors.black, size: 20,),
),
),
),
],
),
automaticallyImplyLeading: false,
flexibleSpace: FlexibleSpaceBar(
collapseMode: CollapseMode.pin,
background: Container(color: Colors.grey,
child: Stack(
children: [
PageView.builder(
controller: _controller,
itemCount: widget.hub.medias.length,
itemBuilder: (context, index) {
Url postMedia = widget.hub.medias[index];
return GestureDetector(
onTap: () => push(
context,
FullScreenImageViewer(
imageUrl: postMedia.url)),
child: displayPostImage(postMedia.url));
}),
if (widget.hub.medias.length > 1)
Padding(
padding: const EdgeInsets.only(bottom: 30.0),
child: Align(
alignment: Alignment.bottomCenter,
child: SmoothPageIndicator(
controller: _controller,
count: widget.hub.medias.length,
effect: ScrollingDotsEffect(
dotWidth: 6,
dotHeight: 6,
dotColor: isDarkMode(context)
? Colors.white54
: Colors.black54,
activeDotColor: Color(COLOR_PRIMARY)),
),
),
),
],
),
)
),
),
_service.isLoading?
SliverFillRemaining(
child: Center(
child: CircularProgressIndicator(),
),
):
SliverList(
delegate: SliverChildListDelegate([
if(widget.hub.userId != MyAppState.currentUser!.userID)
_isSubLoading?
Center(
child: Padding(
padding: EdgeInsets.all(5),
child: CircularProgressIndicator(),
),
):
Padding(
padding: EdgeInsets.symmetric(horizontal: 5),
child: widget.hub.shareUserIds.contains(MyAppState.currentUser!.userID)?
ElevatedButton(
onPressed: () async {
setState(() {
_isSubLoading = true;
});
await Provider.of<HubService>(context, listen: false).unsubscribe(widget.hub);
setState(() {
_isSubLoading = false;
widget.hub.shareUserIds.remove(MyAppState.currentUser!.userID);
});
},
style: ElevatedButton.styleFrom(
primary: Colors.red
),
child: Text(
"Unsubscribe",
),
):
ElevatedButton(
onPressed: () async {
setState(() {
_isSubLoading = true;
});
await Provider.of<HubService>(context, listen: false).subscribe(widget.hub);
setState(() {
_isSubLoading = false;
widget.hub.shareUserIds.add(MyAppState.currentUser!.userID);
});
},
style: ElevatedButton.styleFrom(
primary: Colors.green
),
child: Text(
"Subscribe",
),
),
),
Padding(
padding: EdgeInsets.all(15,),
child: Text(
widget.hub.name,
style: TextStyle(
color: Colors.black,
fontSize: 18,
fontWeight: FontWeight.bold
),
),
),
..._service.hubResults.map((e) {
if(e.isAuction && (e.auctionEnded || DateTime.now().isAfter(e.auctionEndTime??DateTime.now()))) {
return Container();
}
return PostWidget(post: e);
}).toList(),
if(_service.noResult)
Padding(
padding: EdgeInsets.all(20),
child: Text(
'No results for this hub',
style: TextStyle(
fontSize: 18,
fontWeight: FontWeight.bold
),
),
),
if(_service.isMore)
Center(
child: Container(
padding: EdgeInsets.all(5),
child: CircularProgressIndicator(),
),
)
]),
)
],
),
)
);
}
}
You can try using 2 lambdas and S3. These resources are very cheap and you will only be charged once the app has extreme usage ( if the business model is good then high usage -> higher income).
The first lambda will be used to push a text-document mapping to an S3 json file.
the second lambda will basically be your search api, you will use it to query the JSON in s3 and return the results.
The drawback will probably be the latency from s3 to lambda.
I use this with Vue js
query(collection(db,'collection'),where("name",">=",'searchTerm'),where("name","<=","~"))
I also couldn't manage to create a search function to Firebase using the suggestions and Firebase tools so I created my own "field-string contains search-string(substring) check", using the .contains() Kotlin function:
firestoreDB.collection("products")
.get().addOnCompleteListener { task->
if (task.isSuccessful){
val document = task.result
if (!document.isEmpty) {
if (document != null) {
for (documents in document) {
var name = documents.getString("name")
var type = documents.getString("type")
if (name != null && type != null) {
if (name.contains(text, ignoreCase = true) || type.contains(text, ignoreCase = true)) {
// do whatever you want with the document
} else {
showNoProductsMsg()
}
}
}
}
binding.progressBarSearch.visibility = View.INVISIBLE
} else {
showNoProductsMsg()
}
} else{
showNoProductsMsg()
}
}
First, you get ALL the documents in the collection you want, then you filter them using:
for (documents in document) {
var name = documents.getString("name")
var type = documents.getString("type")
if (name != null && type != null) {
if (name.contains(text, ignoreCase = true) || type.contains(text, ignoreCase = true)) {
//do whatever you want with this document
} else {
showNoProductsMsg()
}
}
}
In my case, I filtered them all by the name of the product and its type, then I used the boolean name.contains(string, ignoreCase = true) OR type.contains(string, ignoreCase = true, string is the text I got in the search bar of my app and I recommend you to use ignoreCase = true. With this setence being true, you can do whatever you want with the document.
I guess this is the best workaround since Firestore only supports number and exacts strings queries, so if your code didn't work doing this:
collection.whereGreaterThanOrEqualTo("name", querySearch)
collection.whereLessThanOrEqualTo("name", querySearch)
You're welcome :) because what I did works!
Firebase suggests Algolia or ElasticSearch for Full-Text search, but a cheaper alternative might be MongoDB. The cheapest cluster (approx US$10/mth) allows you to index for full-text.
We can use the back-tick to print out the value of a string. This should work:
where('name', '==', `${searchTerm}`)
So I have this issue where every time I add a new user account, it kicks out the current user that is already signed in. I read the firebase api and it said that "If the new account was created, the user is signed in automatically" But they never said anything else about avoiding that.
//ADD EMPLOYEES
addEmployees: function(formData){
firebase.auth().createUserWithEmailAndPassword(formData.email, formData.password).then(function(data){
console.log(data);
});
},
I'm the admin and I'm adding accounts into my site. I would like it if I can add an account without being signed out and signed into the new account. Any way i can avoid this?
Update 20161110 - original answer below
Also, check out this answer for a different approach.
Original answer
This is actually possible.
But not directly, the way to do it is to create a second auth reference and use that to create users:
var config = {apiKey: "apiKey",
authDomain: "projectId.firebaseapp.com",
databaseURL: "https://databaseName.firebaseio.com"};
var secondaryApp = firebase.initializeApp(config, "Secondary");
secondaryApp.auth().createUserWithEmailAndPassword(em, pwd).then(function(firebaseUser) {
console.log("User " + firebaseUser.uid + " created successfully!");
//I don't know if the next statement is necessary
secondaryApp.auth().signOut();
});
If you don't specify which firebase connection you use for an operation it will use the first one by default.
Source for multiple app references.
EDIT
For the actual creation of a new user, it doesn't matter that there is nobody or someone else than the admin, authenticated on the second auth reference because for creating an account all you need is the auth reference itself.
The following hasn't been tested but it is something to think about
The thing you do have to think about is writing data to firebase. Common practice is that users can edit/update their own user info so when you use the second auth reference for writing this should work. But if you have something like roles or permissions for that user make sure you write that with the auth reference that has the right permissions. In this case, the main auth is the admin and the second auth is the newly created user.
Update 20161108 - original answer below
Firebase just released its firebase-admin SDK, which allows server-side code for this and other common administrative use-cases. Read the installation instructions and then dive into the documentation on creating users.
original answer
This is currently not possible. Creating an Email+Password user automatically signs that new user in.
I just created a Firebase Function that triggers when a Firestore document is Created (with rules write-only to admin user). Then use admin.auth().createUser() to create the new user properly.
export const createUser = functions.firestore
.document('newUsers/{userId}')
.onCreate(async (snap, context) => {
const userId = context.params.userId;
const newUser = await admin.auth().createUser({
disabled: false,
displayName: snap.get('displayName'),
email: snap.get('email'),
password: snap.get('password'),
phoneNumber: snap.get('phoneNumber')
});
// You can also store the new user in another collection with extra fields
await admin.firestore().collection('users').doc(newUser.uid).set({
uid: newUser.uid,
email: newUser.email,
name: newUser.displayName,
phoneNumber: newUser.phoneNumber,
otherfield: snap.get('otherfield'),
anotherfield: snap.get('anotherfield')
});
// Delete the temp document
return admin.firestore().collection('newUsers').doc(userId).delete();
});
You can Algo use functions.https.onCall()
exports.createUser= functions.https.onCall((data, context) => {
const uid = context.auth.uid; // Authorize as you want
// ... do the same logic as above
});
calling it.
const createUser = firebase.functions().httpsCallable('createUser');
createUser({userData: data}).then(result => {
// success or error handling
});
Swift 5: Simple Solution
First store the current user in a variable called originalUser
let originalUser = Auth.auth().currentUser
Then, in the completion handler of creating a new user, use the updateCurrentUser method to restore the original user
Auth.auth().updateCurrentUser(originalUser, completion: nil)
Here is a simple solution using web SDKs.
Create a cloud function (https://firebase.google.com/docs/functions)
import admin from 'firebase-admin';
import * as functions from 'firebase-functions';
const createUser = functions.https.onCall((data) => {
return admin.auth().createUser(data)
.catch((error) => {
throw new functions.https.HttpsError('internal', error.message)
});
});
export default createUser;
Call this function from your app
import firebase from 'firebase/app';
const createUser = firebase.functions().httpsCallable('createUser');
createUser({ email, password })
.then(console.log)
.catch(console.error);
Optionally, you can set user document information using the returned uid.
createUser({ email, password })
.then(({ data: user }) => {
return database
.collection('users')
.doc(user.uid)
.set({
firstname,
lastname,
created: new Date(),
});
})
.then(console.log)
.catch(console.error);
I got André's very clever workaround working in Objective-C using the Firebase iOS SDK:
NSString *plistPath = [[NSBundle mainBundle] pathForResource:#"GoogleService-Info" ofType:#"plist"];
FIROptions *secondaryAppOptions = [[FIROptions alloc] initWithContentsOfFile:plistPath];
[FIRApp configureWithName:#"Secondary" options:secondaryAppOptions];
FIRApp *secondaryApp = [FIRApp appNamed:#"Secondary"];
FIRAuth *secondaryAppAuth = [FIRAuth authWithApp:secondaryApp];
[secondaryAppAuth createUserWithEmail:user.email
password:user.password
completion:^(FIRUser * _Nullable user, NSError * _Nullable error) {
[secondaryAppAuth signOut:nil];
}];
Update for Swift 4
I have tried a few different options to create multiple users from a single account, but this is by far the best and easiest solution.
Original answer by Nico
First Configure firebase in your AppDelegate.swift file
func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -> Bool {
// Override point for customization after application launch.
FirebaseApp.configure()
FirebaseApp.configure(name: "CreatingUsersApp", options: FirebaseApp.app()!.options)
return true
}
Add the following code to action where you are creating the accounts.
if let secondaryApp = FirebaseApp.app(name: "CreatingUsersApp") {
let secondaryAppAuth = Auth.auth(app: secondaryApp)
// Create user in secondary app.
secondaryAppAuth.createUser(withEmail: email, password: password) { (user, error) in
if error != nil {
print(error!)
} else {
//Print created users email.
print(user!.email!)
//Print current logged in users email.
print(Auth.auth().currentUser?.email ?? "default")
try! secondaryAppAuth.signOut()
}
}
}
}
You can use firebase function for add users.
const functions = require('firebase-functions');
const admin = require('firebase-admin');
admin.initializeApp();
const cors = require('cors')({
origin: true,
});
exports.AddUser = functions.https.onRequest(( req, res ) => {
// Grab the text parameter.
cors( req, res, () => {
let email = req.body.email;
let passwd = req.body.passwd;
let role = req.body.role;
const token = req.get('Authorization').split('Bearer ')[1];
admin.auth().verifyIdToken(token)
.then(
(decoded) => {
// return res.status(200).send( decoded )
return creatUser(decoded);
})
.catch((err) => {
return res.status(401).send(err)
});
function creatUser(user){
admin.auth().createUser({
email: email,
emailVerified: false,
password: passwd,
disabled: false
})
.then((result) => {
console.log('result',result);
return res.status(200).send(result);
}).catch((error) => {
console.log(error.message);
return res.status(400).send(error.message);
})
}
});
});
CreateUser(){
//console.log('Create User')
this.submitted = true;
if (this.myGroup.invalid) {
return;
}
let Email = this.myGroup.value.Email;
let Passwd = this.myGroup.value.Passwd;
let Role = 'myrole';
let TechNum = this.myGroup.value.TechNum;
let user = JSON.parse(localStorage.getItem('user'));
let role = user.role;
let AdminUid = user.uid;
let authToken = user.stsTokenManager.accessToken;
let httpHeaders = new HttpHeaders().set('Authorization', 'Bearer ' + authToken);
let options = { headers: httpHeaders };
let params = { email:Email,passwd:Passwd,role:Role };
this.httpClient.post('https://us-central1-myproject.cloudfunctions.net/AddUser', params, options)
.subscribe( val => {
//console.log('Response from cloud function', val );
let createdUser:any = val;
//console.log(createdUser.uid);
const userRef: AngularFirestoreDocument<any> = this.afs.doc(`users/${createdUser.uid}`);
const userUpdate = {
uid: createdUser.uid,
email: createdUser.email,
displayName: null,
photoURL: null,
emailVerified: createdUser.emailVerified,
role: Role,
TechNum:TechNum,
AccountAccess:this.AccountAccess,
UserStatus:'open',
OwnerUid:AdminUid,
OwnerUidRole:role,
RootAccountAccess:this.RootAccountAccess
}
userRef.set(userUpdate, {
merge: false
});
this.toastr.success('Success, user add','Success');
this.myGroup.reset();
this.submitted = false;
},
err => {
console.log('HTTP Error', err.error)
this.toastr.error(err.error,'Error')
},
() => console.log('HTTP request completed.')
);
}
On the web, this is due to unexpected behavior when you call createUserWithEmailAndPassword out of the registration context; e.g. inviting a new user to your app by creating a new user account.
Seems like, createUserWithEmailAndPassword method triggers a new refresh token and user cookies are updated too. (This side-effect is not documented)
Here is a workaround for Web SDK:
After creating the new user;
firebase.auth().updateCurrentUser (loggedInUser.current)
provided that you initiate loggedInUser with the original user beforehand.
Hey i had similar problem ,trying to create users through admin , as it is not possible to signUp user without signIn ,I created a work around ,adding it below with steps
Instead of signup create a node in firebase realtime db with email as key (firebase do not allow email as key so I have created a function to generate key from email and vice versa, I will attach the functions below)
Save a initial password field while saving user (can even hash it with bcrypt or something, if you prefer though it will be used one time only)
Now Once user try to login check if any node with that email (generate key from email) exist in the db and if so then match the password provided.
If the password matched delete the node and do authSignUpWithEmailandPassword with provided credentials.
User is registered successfully
//Sign In
firebaseDB.child("users").once("value", (snapshot) => {
const users = snapshot.val();
const userKey = emailToKey(data.email);
if (Object.keys(users).find((key) => key === userKey)) {
setError("user already exist");
setTimeout(() => {
setError(false);
}, 2000);
setLoading(false);
} else {
firebaseDB
.child(`users`)
.child(userKey)
.set({ email: data.email, initPassword: data.password })
.then(() => setLoading(false))
.catch(() => {
setLoading(false);
setError("Error in creating user please try again");
setTimeout(() => {
setError(false);
}, 2000);
});
}
});
//Sign Up
signUp = (data, setLoading, setError) => {
auth
.createUserWithEmailAndPassword(data.email, data.password)
.then((res) => {
const userDetails = {
email: res.user.email,
id: res.user.uid,
};
const key = emailToKey(data.email);
app
.database()
.ref(`users/${key}`)
.remove()
.then(() => {
firebaseDB.child("users").child(res.user.uid).set(userDetails);
setLoading(false);
})
.catch(() => {
setLoading(false);
setError("error while registering try again");
setTimeout(() => setError(false), 4000);
});
})
.catch((err) => {
setLoading(false);
setError(err.message);
setTimeout(() => setError(false), 4000);
});
};
//Function to create a valid firebase key from email and vice versa
const emailToKey = (email) => {
//firebase do not allow ".", "#", "$", "[", or "]"
let key = email;
key = key.replace(".", ",0,");
key = key.replace("#", ",1,");
key = key.replace("$", ",2,");
key = key.replace("[", ",3,");
key = key.replace("]", ",4,");
return key;
};
const keyToEmail = (key) => {
let email = key;
email = email.replace(",0,", ".");
email = email.replace(",1,", "#");
email = email.replace(",2,", "$");
email = email.replace(",3,", "[");
email = email.replace(",4,", "]");
return email;
};
If you want to do it in your front end create a second auth reference use it to create other users and sign out and delete that reference. If you do it this way you won't be signed out when creating a new user and you won't get the error that the default firebase app already exists.
const createOtherUser =()=>{
var config = {
//your firebase config
};
let secondaryApp = firebase.initializeApp(config, "secondary");
secondaryApp.auth().createUserWithEmailAndPassword(email, password).then((userCredential) => {
console.log(userCredential.user.uid);
}).then(secondaryApp.auth().signOut()
)
.then(secondaryApp.delete()
)
}
Update 19.05.2022 - using #angular/fire (latest available = v.7.3.0)
If you are not using firebase directly in your app, but use e.g. #angular/fire for auth purposes only, you can use the same approach as suggested earlier as follows with the #angular/fire library:
import { Auth, getAuth, createUserWithEmailAndPassword } from '#angular/fire/auth';
import { deleteApp, initializeApp } from '#angular/fire/app';
import { firebaseConfiguration } from '../config/app.config'; // <-- Your project's configuration here.
const tempApp = initializeApp(firebaseConfiguration, "tempApp");
const tempAppAuth = getAuth(tempApp);
await createUserWithEmailAndPassword(tempAppAuth, email, password)
.then(async (newUser) => {
resolve( () ==> {
// Do something, e.g. add user info to database
});
})
.catch(error => reject(error))
.finally( () => {
tempAppAuth.signOut()
.then( () => deleteApp(tempApp));
});
The Swift version:
FIRApp.configure()
// Creating a second app to create user without logging in
FIRApp.configure(withName: "CreatingUsersApp", options: FIRApp.defaultApp()!.options)
if let secondaryApp = FIRApp(named: "CreatingUsersApp") {
let secondaryAppAuth = FIRAuth(app: secondaryApp)
secondaryAppAuth?.createUser(...)
}
Here is a Swift 3 adaptaion of Jcabrera's answer :
let bundle = Bundle.main
let path = bundle.path(forResource: "GoogleService-Info", ofType: "plist")!
let options = FIROptions.init(contentsOfFile: path)
FIRApp.configure(withName: "Secondary", options: options!)
let secondary_app = FIRApp.init(named: "Secondary")
let second_auth = FIRAuth(app : secondary_app!)
second_auth?.createUser(withEmail: self.username.text!, password: self.password.text!)
{
(user,error) in
print(user!.email!)
print(FIRAuth.auth()?.currentUser?.email ?? "default")
}
If you are using Polymer and Firebase (polymerfire) see this answer: https://stackoverflow.com/a/46698801/1821603
Essentially you create a secondary <firebase-app> to handle the new user registration without affecting the current user.
Android solution (Kotlin):
1.You need FirebaseOptions BUILDER(!) for setting api key, db url, etc., and don't forget to call build() at the end
2.Make a secondary auth variable by calling FirebaseApp.initializeApp()
3.Get instance of FirebaseAuth by passing your newly created secondary auth, and do whatever you want (e.g. createUser)
// 1. you can find these in your project settings under general tab
val firebaseOptionsBuilder = FirebaseOptions.Builder()
firebaseOptionsBuilder.setApiKey("YOUR_API_KEY")
firebaseOptionsBuilder.setDatabaseUrl("YOUR_DATABASE_URL")
firebaseOptionsBuilder.setProjectId("YOUR_PROJECT_ID")
firebaseOptionsBuilder.setApplicationId("YOUR_APPLICATION_ID") //not sure if this one is needed
val firebaseOptions = firebaseOptionsBuilder.build()
// indeterminate progress dialog *ANKO*
val progressDialog = indeterminateProgressDialog(resources.getString(R.string.progressDialog_message_registering))
progressDialog.show()
// 2. second auth created by passing the context, firebase options and a string for secondary db name
val newAuth = FirebaseApp.initializeApp(this#ListActivity, firebaseOptions, Constants.secondary_db_auth)
// 3. calling the create method on our newly created auth, passed in getInstance
FirebaseAuth.getInstance(newAuth).createUserWithEmailAndPassword(email!!, password!!)
.addOnCompleteListener { it ->
if (it.isSuccessful) {
// 'it' is a Task<AuthResult>, so we can get our newly created user from result
val newUser = it.result.user
// store wanted values on your user model, e.g. email, name, phonenumber, etc.
val user = User()
user.email = email
user.name = name
user.created = Date().time
user.active = true
user.phone = phone
// set user model on /db_root/users/uid_of_created_user/, or wherever you want depending on your structure
FirebaseDatabase.getInstance().reference.child(Constants.db_users).child(newUser.uid).setValue(user)
// send newly created user email verification link
newUser.sendEmailVerification()
progressDialog.dismiss()
// sign him out
FirebaseAuth.getInstance(newAuth).signOut()
// DELETE SECONDARY AUTH! thanks, Jimmy :D
newAuth.delete()
} else {
progressDialog.dismiss()
try {
throw it.exception!!
// catch exception for already existing user (e-mail)
} catch (e: FirebaseAuthUserCollisionException) {
alert(resources.getString(R.string.exception_FirebaseAuthUserCollision), resources.getString(R.string.alertDialog_title_error)) {
okButton {
isCancelable = false
}
}.show()
}
}
}
For Android, i suggest a simpler way to do it, without having to provide api key, application id...etc by hand by just using the FirebaseOptions of the default instance.
val firebaseDefaultApp = Firebase.auth.app
val signUpAppName = firebaseDefaultApp.name + "_signUp"
val signUpApp = try {
FirebaseApp.initializeApp(
context,
firebaseDefaultApp.options,
signUpAppName
)
} catch (e: IllegalStateException) {
// IllegalStateException is throw if an app with the same name has already been initialized.
FirebaseApp.getInstance(signUpAppName)
}
// Here is the instance you can use to sign up without triggering auth state on the default Firebase.auth
val signUpFirebaseAuth = Firebase.auth(signUpApp)
How to use ?
signUpFirebaseAuth
.createUserWithEmailAndPassword(email, password)
.addOnSuccessListener {
// Optional, you can send verification email here if you need
// As soon as the sign up with sign in is over, we can sign out the current user
firebaseAuthSignUp.signOut()
}
.addOnFailureListener {
// Log
}
My solution to this question is to store the User Name/Email and password in a static class and then add a new user log out the new user and immediately log in as the admin user(id pass you saved). Works like a charm for me :D
This is a version for Kotlin:
fun createUser(mail: String, password: String) {
val opts = FirebaseOptions.fromResource(requireContext())
if (opts == null) return
val app = Firebase.initialize(requireContext(), opts, "Secondary")
FirebaseAuth.getInstance(app)
.createUserWithEmailAndPassword(mail, password)
.addOnSuccessListener {
app.delete()
doWhateverWithAccount(it)
}.addOnFailureListener {
app.delete()
showException(it)
}
}
It uses the configuration from your default Firebase application instance, just under a different name.
It also deletes the newly created instance afterwards, so you can call this multiple times without any exception about already existing Secondary application.