Develop Reference

Unable to retrieve data from POST request

I am working in JAVA 1.8 to write and using Apache Tomcat to run the server, I am unable to retrieve data from a POST request i.e in JSON.
I actually need it in an HashMap and I can even parse and convert it into HashMap even if it is readable in JSON. I have tried several links on the internet and I always get exception like Could not deserialize to type interface PACKAGE NAME.
#POST
#Produces("application/json")
#Consumes("application/json")
#Path("ClassifyCase")
public Rules Classify(HttpServletRequest request) {
StringBuffer jb = new StringBuffer();
String line = null;
try {
BufferedReader reader = request.getReader();
while ((line = reader.readLine()) != null)
jb.append(line);
} catch (Exception e) { System.out.println("Buffer Reader Error"); }
System.out.println("What I read: "+jb);
System.out.println("Here la la l ala ");
// System.out.println("Case: ++ "+Case.toString());
System.out.println("Here la la l ala ");
Rules foundRule = new Rules();
// List<Rules> objListRules = new ArrayList<Rules>();
try
{
DataAccessInterface objDAInterface = new RuleDataAdapter();
AbstractDataBridge objADBridge = new DatabaseStorage(objDAInterface);
// foundRule = objADBridge.Classify(Case);
logger.info("Classification done!");
}
catch(Exception ex)
{
logger.info("Error in classification");
System.out.println("Couldnt Classify Properly!");
// return
}
return foundRule;
}
Can someone please share a guide on how can I receive this data and convert it into a Map or either I can directly get a Map!

I strongly recommend you to use this library of JSON..
You can find it in Maven Repository and it's so easy to parse a JSON to a Map or to a JSONArray or JSONObject... depends of your necessity what you want to do..
Here is a example show how to parse a JSON to a HashMap
Map<String, Object> map = new JSONObject(--JSONString here--).toMap();
And that's all...
Now, if your JSON has a list of objects, i mean like a list of maps, what you just need to do is this...
JSONArray jsonArray = new JSONArray(--JSON string here--);
for(int i = 0; i < jsonArray.length(); i++){
Map<String, Object> map = jsonArray.getJSONObject(i).toMap();
}
Here is the explanation.
You take you JSON string and pass it as a parameter to the JSONArray,what JSONArray does is, take your json string a parse it to like a list
Then you make a for to get each Object of that list and parse it to a map.
Note: what the JSONObject does, is take the object of the JSONArray and parse it... you can parse it to a map or you can get each object of that map..

String jsonString = "{\n" +
"\t\"1\": \"1\",\n" +
"\t\"FPG\": \"50\",\n" +
"\t\"Symptoms\": \"Yes\"\n" +
"}";
Map<String, String> map = new Gson().fromJson(jsonString, Map.class);
for (String key: map.keySet()) {
System.out.println(map.get(key));
}

The request you send does not contain proper JSON in the body. You are missing the commas ",". It should be something like this:
{
"1":"1",
"FPG":"50",
"Symptoms":"yes"
}
Just change it and give proper JSON format to the message.
Even if the request was not in your control, I would strongly suggest that you contacted the service that creates the message and asked from them to fix it.
It would be the last resort for me to make my own deserializer to handle an "inproper" message.
An easy way to check if your JSON is properly formated is an online formatter, e.g. https://jsonformatter.org/

From Angular Actual parameter value is "Ébénisterie" but in JAVA getting value "Ã?bénisterie"

From Angular, there is one parameter and the value of that parameter is Ébénisterie but when I print the value of that variable in java then I got Ã?bénisterie can you please let me know how I can convert it to original text Ébénisterie? Which Encode/decode I have to apply?
I have tried the following thing.
new String(readable.getBytes("ISO-8859-15"), "UTF-8");
new String(readable.getBytes("UTF-8"), "ISO-8859-15");
but it's not working.
String readable ="�bénisterie Distinction";
String test = null;
try {
test = new String(readable.getBytes("ISO-8859-15"), "UTF-8");
System.out.println("test"+test);
} catch (UnsupportedEncodingException e) {
}
Expected: Ébénisterie
Actual: �bénisterie

After long research didn't find anything.
So got one solution in mind that BASE64 Encode decode so now from Angularjs sending encoded text and In java side, I have decoded the text.
Here, is the sample code
Angularjs
window.btoa("Ébénisterie")
JAVA
String actualString= new String(Base64.getDecoder().decode("ENCODED STRING"));

After converting an object to a String, how can I get the object back from the string again?

I was creating a simple android application in which I am converting an object to String. How can I re-convert the object from the string?
I am converting my object to String using the following line of code.
String convertedString = object.toString();

You can't**, because that is not what the toString method is for. It's used to make a readable representation of your Object, but it's not meant for saving and later reloading.
What you are looking for instead is Serialization. See this tutorial here to get started:
http://www.tutorialspoint.com/java/java_serialization.htm
** Technically you could, but you shouldn't.

You can do it in two ways:
Java Serialization
Using Gson library (more simple), remember the the purpose of this lib is to convert simply json to object and viceversa when working with REST services.
Hope it helps

You can use Serialization to convert object to string and vise versa:
String serializedObject = "";
// serialize the object
try {
ByteArrayOutputStream bo = new ByteArrayOutputStream();
ObjectOutputStream so = new ObjectOutputStream(bo);
so.writeObject(myObject);
so.flush();
serializedObject = bo.toString();
} catch (Exception e) {
System.out.println(e);
}
// deserialize the object
try {
byte b[] = serializedObject.getBytes();
ByteArrayInputStream bi = new ByteArrayInputStream(b);
ObjectInputStream si = new ObjectInputStream(bi);
MyObject obj = (MyObject) si.readObject();
} catch (Exception e) {
System.out.println(e);
}

Use Java Serialization for doing same.
Go with below link for better understand how to convert java object.
Ex. http://www.geeksforgeeks.org/serialization-in-java/
Also You can go with this link:
How to convert the following json string to java object?

Unable to parse JSON from url

Write a piece of code that will query a URL that returns JSON and can parse the JSON string to pull out pieces of information. The information that should be parsed and returned is the pageid and the list of “See Also” links. Those links should be formatted to be actual links that can be used by a person to find the appropriate article.
Use the Wikipedia API for the query. A sample query is:
URL
Other queries can be generated changing the “titles” portion of the query string. The code to parse the JSON and pull the “See Also” links should be generic enough to work on any Wikipedia article.
I tried writing the below code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.URL;
import org.json.JSONException;
import org.json.JSONObject;
public class JsonRead {
private static String readUrl(String urlString) throws Exception {
BufferedReader reader = null;
try {
URL url = new URL(urlString);
reader = new BufferedReader(new InputStreamReader(url.openStream()));
StringBuffer buffer = new StringBuffer();
int read;
char[] chars = new char[1024];
while ((read = reader.read(chars)) != -1)
buffer.append(chars, 0, read);
return buffer.toString();
} finally {
if (reader != null)
reader.close();
}
}
public static void main(String[] args) throws IOException, JSONException {
JSONObject json;
try {
json = new JSONObject(readUrl("https://en.wikipedia.org/w/api.php?format=json&action=query&titles=SMALL&prop=revisions&rvprop=content"));
System.out.println(json.toString());
System.out.println(json.get("pageid"));
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
I have used the json jar from the below link in eclipse:
Json jar
When I run the above code I am getting the below error;
org.json.JSONException: JSONObject["pageid"] not found.
at org.json.JSONObject.get(JSONObject.java:471)
at JsonRead.main(JsonRead.java:35)
How can I extract the details of the pageid and also the "See Also" links from the url?
I have never worked on JSON before hence kindly let me know how to proceed here
The json:
{
"batchcomplete":"",
"query":{
"pages":{
"1808130":{
"pageid":1808130,
"ns":0,
"title":"SMALL",
"revisions":[
{
"contentformat":"text/x-wiki",
"contentmodel":"wikitext",
"*":"{{About|the ALGOL-like programming language|the scripting language formerly named Small|Pawn (scripting language)}}\n\n'''SMALL''', Small Machine Algol Like Language, is a [[computer programming|programming]] [[programming language|language]] developed by Dr. [[Nevil Brownlee]] of [[Auckland University]].\n\n==History==\nThe aim of the language was to enable people to write [[ALGOL]]-like code that ran on a small machine. It also included the '''string''' type for easier text manipulation.\n\nSMALL was used extensively from about 1980 to 1985 at [[Auckland University]] as a programming teaching aid, and for some internal projects. Originally written to run on a [[Burroughs Corporation]] B6700 [[Main frame]] in [[Fortran]] IV, subsequently rewritten in SMALL and ported to a DEC [[PDP-10]] Architecture (on the [[Operating System]] [[TOPS-10]]) and IBM S360 Architecture (on the Operating System VM/[[Conversational Monitor System|CMS]]).\n\nAbout 1985, SMALL had some [[Object-oriented programming|object-oriented]] features added to handle structures (that were missing from the early language), and to formalise file manipulation operations.\n\n==See also==\n*[[ALGOL]]\n*[[Lua (programming language)]]\n*[[Squirrel (programming language)]]\n\n==References==\n*[http://www.caida.org/home/seniorstaff/nevil.xml Nevil Brownlee]\n\n[[Category:Algol programming language family]]\n[[Category:Systems programming languages]]\n[[Category:Procedural programming languages]]\n[[Category:Object-oriented programming languages]]\n[[Category:Programming languages created in the 1980s]]"
}
]
}
}
}
}

If You Read your Exception Carefully you will find your solution at your own.
Exception in thread "main" org.json.JSONException: A JSONObject text must begin with '{' at 1 [character 2 line 1]
at org.json.JSONTokener.syntaxError(JSONTokener.java:433)
Your Exception says A JSONObject text must begin with '{' it means the the json you received from the api is probably not Correct.
So, I suggest you to debug your code and try to find out what you actually received in your String Variable jsonText.

You get the exception org.json.JSONException: JSONObject["pageid"] not found. when calling json.get("pageid") because pageid is not a direct sub-element of your root. You have to go all the way down through the object graph:
int pid = json.getJSONObject("query")
.getJSONObject("pages")
.getJSONObject("1808130")
.getInt("pageid");
If you have an array in there you will even have to iterate the array elements (or pick the one you want).
Edit Here's the code to get the field containing the 'see also' values
String s = json.getJSONObject("query")
.getJSONObject("pages")
.getJSONObject("1808130")
.getJSONArray("revisions")
.getJSONObject(0)
.getString("*");
The resulting string contains no valid JSON. You will have to parse it manually.

Faster alternative to JsonPath in Java

JsonPath seems to be pretty slow for large JSON files.
In my project, I'd like a user to be able to pass an entire query as a string. I used JsonPath because it lets you do an entire query like $.store.book[3].price all at once by doing JsonPath.read(fileOrString, "$.store.book[3].price", new Filter[0]). Is there a faster method to interact with JSON files in Javascript? It would be ideal to be able to pass the entire query as a string, but I'll write a parser if I have to. Any ideas?
Even small optimizations would be helpful. For instance, I'm currently reading from a JSON file every time I query. Would it be better just to copy the entire file into a string at the beginning and query to the string instead?
EDIT: To those of you saying "this is Javascript, not Java", well, it actually is Java. JsonPath is a Javascript-like query language, but the file I am writing is most assuredly Java. Only the query is written in Javascript. Here's some info about JsonPath, and a snippet of code: https://code.google.com/p/json-path/
List toRet;
String query = "$.store.book[3].price";
try {
// if output is a list, good
toRet = (List) JsonPath.read(filestring_, query, new Filter[0]);
} catch (ClassCastException cce) {
// if output isn't a list, put it in a list
Object outObj = null;
try {
outObj = JsonPath.read(filestring_, query, new Filter[0]);
} catch (Exception e) {
throw new DataSourceException("Invalid file!\n", e, DataSourceException.UNKNOWN);
}