Related
{
"_id" : ObjectId("dde3431134247d401b1cef"),
"_resourceId" : "fwf4-fefre4-ffdfwsc",
"organizationId" : 343203,
"domains" : [
{
"_resourceId" : "da7-cwcwe-2432d",
"name" : "d12.net",
"tenantId" : "A1650",
"application" : "TEST",
"activeInd" : true,
"subdomains" : [
{
"_resourceId" : "fw243-weded3-2eddas",
"name" : "name1",
"clientName" : "Andrew",
"phoneNumber" : "8573458456",
"email" : "modalwindow#gmail.com",
},
{
"_resourceId" : "bce3-cwdd32ede-23ede",
"name" : "name2",
"clientName" : "client2",
"phoneNumber" : "9999999999",
"email" : "test#gmail.com",
}
]
}
]
}
I am using Springboot and MongoTemplate for the find Query. If I want to retrieve a subdomain based on "domains.subdomain.name" field, can I get only the subdomain json from a mongo query, or do I get the entire document and then iterate and filter the subdomain in my java code.
Use $unwind and then $replaceWith
db.collection.aggregate([
{
$match: {
"domains.subdomains.name": "name1"
}
},
{
$unwind: "$domains"
},
{
$unwind: "$domains.subdomains"
},
{
$match: {
"domains.subdomains.name": "name1"
}
},
{
$replaceWith: "$domains.subdomains"
}
])
mongoplayground
I have a following mongoDB document structure -
db.menus.findOne()
{
"_id" : ObjectId("5cf25412326c3f4f26df039b"),
"restaurantId" : "301728",
"items" : [
{
"itemId" : "CEBM4H41JR",
"name" : "Crun Chicken",
"imageUrl" : "",
"price" : 572,
"attributes" : [
"Tasty",
"Spicy"
]
},
{
"itemId" : "53Q0XS3HPR",
"name" : "Devils Chicken",
"imageUrl" : "",
"price" : 595,
"attributes" : [
"Gravy",
"Salty"
]
}
]
}
I am trying to write a query to get all the menus based on the "attributes" field under "items" in the document.
I have done the following to get the menus if "name" of "items" is given and I am getting a result -
db.menus.find({ 'items' : {$elemMatch : {'name' : {$regex : "Chicken Thali", $options: 'i' }}}}).pretty()
I have tried this for getting the result for attributes but this is not working -
db.menus.find({'items' : {$elemMatch : {'attributes' : {$all : [{$regex : "Tasty", $options: 'i' }]}}}})
How do I get the list and I also want to write this query for mongoRepository in a spring boot application?
Further, based on the restaurantId's obtained, I have to query restaurant collection in order to find all the restaurants in restaurants collection having the following structure -
{
"_id" : ObjectId("5cf2540e326c3f4f26de93dd"),
"restaurantId" : "301728",
"name" : "Desire Foods",
"imageUrl" : "https://b.zmtcdn.com/data/pictures/8/301728/d690ccb500d746530f56e1d637949da2_featured_v2.jpg",
"latitude" : 28.4900591,
"longitude" : 77.3066401,
"attributes" : [
"Chinese",
" Fast Food",
" Bakery"
],
"opensAt" : "09:30",
"closesAt" : "22:30"
}
Is the whole operation possible in a single query?
I think you can modify your query to use $in instead of $all.
To achieve your intended result, you can try:
db.collection.aggregate([
{
"$match": {
"items": {
"$elemMatch": {
"attributes": {
"$in": [
"Tasty"
]
}
}
}
}
},
{
"$lookup": {
"from": "restaurant",
"localField": "restaurantId",
"foreignField": "restaurantId",
"as": "restaurants"
}
},
{
"$unwind": "restaurants"
},
{
"$replaceRoot": { "newRoot": "$restaurants" }
}
])
Use $match at appropriate stages as needed to limit the documents pulled in memory
I'm working with documents that contain music playlists.
Each document has this structure:
{
"user_id": "5858",
"playlists": [
{
"name": "My Playlist",
"guild_ids": ["7575"],
"items": [
{
"title": "title",
"url": "url",
"duration": 200000
}
]
}
]
}
I would like to extract all playlists from the same guild.
But the thing is that i'd like the results to be returned in a single document. One single document with a list of playlists.
The expected result for guild_id=5656 would be like this:
{
"playlists": [
{
"name": "My Playlist",
"guild_ids": ["5656"],
"items": [
{
"title": "title",
"url": "url",
"duration": 200000
}
]
},
// other playlists where guild_ids contains "5656"
]
}
I tried to use aggregation but i always get the same number of documents as the number of unique user_ids. I get the playlists grouped by user_id.
The following query can get us the expected output:
db.collection.aggregate([
{
$unwind:"$playlists"
},
{
$match:{
"playlists.guild_ids":{
$in:["7575"]
}
}
},
{
$group:{
"_id":null,
"playlists":{
$push: "$playlists"
}
}
},
{
$project:{
"_id":0
}
}
]).pretty()
Data set:
{
"_id" : ObjectId("5d88225e38db7cf8d3f75cd6"),
"user_id" : "5858",
"playlists" : [
{
"name" : "My Playlist",
"guild_ids" : [
"7575"
],
"items" : [
{
"title" : "title",
"url" : "url",
"duration" : 200000
}
]
}
]
}
{
"_id" : ObjectId("5d88225e38db7cf8d3f75cd7"),
"user_id" : "5858",
"playlists" : [
{
"name" : "My Playlist 2",
"guild_ids" : [
"1234"
],
"items" : [
{
"title" : "title",
"url" : "url",
"duration" : 200000
}
]
}
]
}
{
"_id" : ObjectId("5d88225e38db7cf8d3f75cd8"),
"user_id" : "5858",
"playlists" : [
{
"name" : "My Playlist 3",
"guild_ids" : [
"7575"
],
"items" : [
{
"title" : "title",
"url" : "url",
"duration" : 200000
}
]
}
]
}
Output:
{
"playlists" : [
{
"name" : "My Playlist",
"guild_ids" : [
"7575"
],
"items" : [
{
"title" : "title",
"url" : "url",
"duration" : 200000
}
]
},
{
"name" : "My Playlist 3",
"guild_ids" : [
"7575"
],
"items" : [
{
"title" : "title",
"url" : "url",
"duration" : 200000
}
]
}
]
}
Query analysis: We are unwinding the playlists, filtering only those which has 7575 guild ID and then grouping them all again.
I'm new in mongo and use mongodb aggregation framework for my queries. I need to retrieve some records which satisfy certain conditions(include pagination+sorting) and also get total count of records.
Now, I perform next steps:
Create $match operator
{ "$match" : { "year" : "2012" , "author.authorName" : { "$regex" :
"au" , "$options" : "i"}}}
Added sorting and pagination
{ "$sort" : { "some_field" : -1}} , { "$limit" : 10} , { "$skip" : 0}
After querying I receive the expected result: 10 documents with all fields.
For pagination I need to know the total count of records which satisfy these conditions, in my case 25.
I use next query to get count : { "$match" : { "year" : "2012" , "author.authorName" : { "$regex" : "au" , "$options" : "i"}}} , { "$group" : { "_id" : "$all" , "reviewsCount" : { "$sum" : 1}}} , { "$sort" : { "some_field" : -1}} , { "$limit" : 10} , { "$skip" : 0}
But I don't want to perform two separate queries: one for retrieving documents and second for total counts of records which satisfy certain conditions.
I want do it in one single query and get result in next format:
{
"result" : [
{
"my_documets": [
{
"_id" : ObjectId("512f1f47a411dc06281d98c0"),
"author" : {
"authorName" : "author name1",
"email" : "email1#email.com"
}
},
{
"_id" : ObjectId("512f1f47a411dc06281d98c0"),
"author" : {
"authorName" : "author name2",
"email" : "email2#email.com"
}
}, .......
],
"total" : 25
}
],
"ok" : 1
}
I tried modify the group operator : { "$group" : { "_id" : "$all" , "author" : "$author" "reviewsCount" : { "$sum" : 1}}}
But in this case I got : "exception: the group aggregate field 'author' must be defined as an expression inside an object". If add all fields in _id then reviewsCount always = 1 because all records are different.
Nobody know how it can be implement in single query ? Maybe mongodb has some features or operators for this case? Implementation with using two separate query reduces performance for querying thousand or millions records. In my application it's very critical performance issue.
I've been working on this all day and haven't been able to find a solution, so thought i'd turn to the stackoverflow community.
Thanks.
You can try using $facet in the aggregation pipeline as
db.name.aggregate([
{$match:{your match criteria}},
{$facet: {
data: [{$sort: sort},{$skip:skip},{$limit: limit}],
count:[{$group: {_id: null, count: {$sum: 1}}}]
}}
])
In data, you'll get your list with pagination and in the count, count variable will have a total count of matched documents.
Ok, I have one example, but I think it's really crazy query, I put it only for fun, but if this example faster than 2 query, tell us about it in the comments please.
For this question i create collection called "so", and put into this collection 25 documents like this:
{
"_id" : ObjectId("512fa86cd99d0adda2a744cd"),
"authorName" : "author name1",
"email" : "email1#email.com",
"c" : 1
}
My query use aggregation framework:
db.so.aggregate([
{ $group:
{
_id: 1,
collection: { $push : { "_id": "$_id", "authorName": "$authorName", "email": "$email", "c": "$c" } },
count: { $sum: 1 }
}
},
{ $unwind:
"$collection"
},
{ $project:
{ "_id": "$collection._id", "authorName": "$collection.authorName", "email": "$collection.email", "c": "$collection.c", "count": "$count" }
},
{ $match:
{ c: { $lte: 10 } }
},
{ $sort :
{ c: -1 }
},
{ $skip:
2
},
{ $limit:
3
},
{ $group:
{
_id: "$count",
my_documets: {
$push: {"_id": "$_id", "authorName":"$authorName", "email":"$email", "c":"$c" }
}
}
},
{ $project:
{ "_id": 0, "my_documets": "$my_documets", "total": "$_id" }
}
])
Result for this query:
{
"result" : [
{
"my_documets" : [
{
"_id" : ObjectId("512fa900d99d0adda2a744d4"),
"authorName" : "author name8",
"email" : "email8#email.com",
"c" : 8
},
{
"_id" : ObjectId("512fa900d99d0adda2a744d3"),
"authorName" : "author name7",
"email" : "email7#email.com",
"c" : 7
},
{
"_id" : ObjectId("512fa900d99d0adda2a744d2"),
"authorName" : "author name6",
"email" : "email6#email.com",
"c" : 6
}
],
"total" : 25
}
],
"ok" : 1
}
By the end, I think that for big collection 2 query (first for data, second for count) works faster. For example, you can count total for collection like this:
db.so.count()
or like this:
db.so.find({},{_id:1}).sort({_id:-1}).count()
I don't fully sure in first example, but in second example we use only cursor, which means higher speed:
db.so.find({},{_id:1}).sort({_id:-1}).explain()
{
"cursor" : "BtreeCursor _id_ reverse",
"isMultiKey" : false,
"n" : 25,
"nscannedObjects" : 25,
"nscanned" : 25,
"nscannedObjectsAllPlans" : 25,
"nscannedAllPlans" : 25,
"scanAndOrder" : false,
!!!!!>>> "indexOnly" : true, <<<!!!!!
"nYields" : 0,
"nChunkSkips" : 0,
"millis" : 0,
...
}
For completeness (full discussion was on the MongoDB Google Groups) here is the aggregation you want:
db.collection.aggregate(db.docs.aggregate( [
{
"$match" : {
"year" : "2012"
}
},
{
"$group" : {
"_id" : null,
"my_documents" : {
"$push" : {
"_id" : "$_id",
"year" : "$year",
"author" : "$author"
}
},
"reviewsCount" : {
"$sum" : 1
}
}
},
{
"$project" : {
"_id" : 0,
"my_documents" : 1,
"total" : "$reviewsCount"
}
}
] )
By the way, you don't need aggregation framework here - you can just use a regular find. You can get count() from a cursor without having to re-query.
I am facing a trouble in the use of ElasticSearch for my java application.
I explain myself, I have a mapping, which is something like :
{
"products": {
"properties": {
"id": {
"type": "long",
"ignore_malformed": false
},
"locations": {
"properties": {
"category": {
"type": "long",
"ignore_malformed": false
},
"subCategory": {
"type": "long",
"ignore_malformed": false
},
"order": {
"type": "long",
"ignore_malformed": false
}
}
},
...
So, as you can see, I receive a list of products, which are composed of locations. In my model, this locations are all the categories' product. It means that a product can be in 1 or more categories. In each of this category, the product has an order, which is the order the client wants to show them.
For instance, a diamond product can have a first place in Jewelry, but the third place in Woman (my examples are not so logic ^^).
So, when I click on Jewelry, I want to show this products, ordered by the field locations.order in this specific category.
For the moment, when I search all the products on a specific category the response for ElasticSearch that I receive is something like :
{"id":5331880,"locations":[{"category":5322606,"order":1},
{"category":5883712,"subCategory":null,"order":3},
{"category":5322605,"subCategory":6032961,"order":2},.......
Is it possible to sort this products, by the element locations.order for the specific category I am searching for ? For instance, if I am querying the category 5322606, I want the order 1 for this product to be taken.
Thank you very much beforehand !
Regards,
Olivier.
First a correction of terminology: in Elasticsearch, "parent/child" refers to completely separate docs, where the child doc points to the parent doc. Parent and children are stored on the same shard, but they can be updated independently.
With your example above, what you are trying to achieve can be done with nested docs.
Currently, your locations field is of type:"object". This means that the values in each location get flattened to look something like this:
{
"locations.category": [5322606, 5883712, 5322605],
"locations.subCategory": [6032961],
"locations.order": [1, 3, 2]
}
In other words, the "sub" fields get flattened into multi-value fields, which is of no use to you, because there is no correlation between category: 5322606 and order: 1.
However, if you change locations to be type:"nested" then internally it will index each location as a separate doc, meaning that each location can be queried independently, using the dedicated nested query and filter.
By default, the nested query will return a _score based upon how well each location matches, but in your case you want to return the highest value of the order field from any matching children. To do this, you'll need to use a custom_score query.
So let's start by creating the index with the appropriate mapping:
curl -XPUT 'http://127.0.0.1:9200/test/?pretty=1' -d '
{
"mappings" : {
"products" : {
"properties" : {
"locations" : {
"type" : "nested",
"properties" : {
"order" : {
"type" : "long"
},
"subCategory" : {
"type" : "long"
},
"category" : {
"type" : "long"
}
}
},
"id" : {
"type" : "long"
}
}
}
}
}
'
The we index your example doc:
curl -XPOST 'http://127.0.0.1:9200/test/products?pretty=1' -d '
{
"locations" : [
{
"order" : 1,
"category" : 5322606
},
{
"order" : 3,
"subCategory" : null,
"category" : 5883712
},
{
"order" : 2,
"subCategory" : 6032961,
"category" : 5322605
}
],
"id" : 5331880
}
'
And now we can search for it using the queries we discussed above:
curl -XGET 'http://127.0.0.1:9200/test/products/_search?pretty=1' -d '
{
"query" : {
"nested" : {
"query" : {
"custom_score" : {
"script" : "doc[\u0027locations.order\u0027].value",
"query" : {
"constant_score" : {
"filter" : {
"and" : [
{
"term" : {
"category" : 5322605
}
},
{
"term" : {
"subCategory" : 6032961
}
}
]
}
}
}
}
},
"score_mode" : "max",
"path" : "locations"
}
}
}
'
Note: the single quotes within the script have been escaped as \u0027 to get around shell quoting. The script actually looks like this: "doc['locations.order'].value"
If you look at the _score from the results, you can see that it has used the order value from the matching location:
{
"hits" : {
"hits" : [
{
"_source" : {
"locations" : [
{
"order" : 1,
"category" : 5322606
},
{
"order" : 3,
"subCategory" : null,
"category" : 5883712
},
{
"order" : 2,
"subCategory" : 6032961,
"category" : 5322605
}
],
"id" : 5331880
},
"_score" : 2,
"_index" : "test",
"_id" : "cXTFUHlGTKi0hKAgUJFcBw",
"_type" : "products"
}
],
"max_score" : 2,
"total" : 1
},
"timed_out" : false,
"_shards" : {
"failed" : 0,
"successful" : 5,
"total" : 5
},
"took" : 9
}
Just add a more updated version related to sorting parent by child field.
We can query parent doc type sorted by child field ('count' e.g.) similar as follows.
https://gist.github.com/robinloxley1/7ea7c4f37a3413b1ca16