org.json.JSONException: A JSONObject text must begin with '{' at character 1

Question

I tried searching for this error. There are many results on google for this search but nothing proved useful to me. This is my web service method

@GET
@Path("/values")
public String test() {

    return "{\"x\":5,\"y\":6}";

}

This is my client code

public class Check {
public static void main(String[] args){
    String url = "http://localhost:8181/math/webapi/values";
    HttpClient httpClient = HttpClientBuilder.create().build();
    HttpGet request = new HttpGet(url);
    try {
        HttpResponse response = httpClient.execute(request);
        String value = response.toString();
        JSONObject json = new JSONObject(value);
        int i = json.getInt("x");
        System.out.println(i);
    }catch (Exception e) {
        e.printStackTrace();
    }       
}

The above code is a starter code and it is for learning how to use it. If this is solved, I have to apply the knowledge in another application. The client side code, I want to use the logic in android.

EDIT

public class Check {
public static void main(String[] args){
    String url = "http://localhost:8181/math/webapi/values";
    HttpClient httpClient = HttpClientBuilder.create().build();
    HttpGet request = new HttpGet(url);
    try {
        HttpResponse response = httpClient.execute(request);
        InputStream value = response.getEntity().getContent();
        BufferedReader br = new BufferedReader(new InputStreamReader(value));
        String jsonValue = br.readLine();
        JSONObject json = new JSONObject(jsonValue);
        int i = json.getInt("x");
        System.out.println(i);
    }catch (Exception e) {
        e.printStackTrace();
    }       
}

Likely your response contains a header? Try printing the value of `value`. **Edit** likely `response.toString` is not overridden from `Object#toString` and you're getting the `ClassName@hashCode` idiom. — Mena, Oct 16 '15 at 13:20
Are you sure the HTTP API you're using is synchronous? Are you sure that `HttpResponse#toString` returns the response rather than the generic `toString`? — T.J. Crowder, Oct 16 '15 at 13:20
Step through the code in a debugger and examine `value`. That's the only rational way to approach figuring this out. If you don't have a debugger, stop what you're doing and get one. — T.J. Crowder, Oct 16 '15 at 13:21
Possible duplicate of [How can I get an http response body as a string in Java?](http://stackoverflow.com/questions/5769717/how-can-i-get-an-http-response-body-as-a-string-in-java) — Thomas Weller, Oct 16 '15 at 13:22
You need to obtain the content of the response and feed it into the JSON parser: http://stackoverflow.com/a/10804506/3215527 — wero, Oct 16 '15 at 13:22
@T.J.Crowder thanks thomas for the suggestion. I'll debug it. — Krishna, Oct 16 '15 at 13:25
@Thomas as I said, it is not the real application. I'm doing it to learn. I have to apply it in another application. — Krishna, Oct 16 '15 at 13:26
@Mena Thanks for the help offered Mena. I'm not getting how can I override toString() using that response in client program. And what you said is right. I got the hashcode value when I did this `System.out.println(response);`. Can you help me further? — Krishna, Oct 16 '15 at 13:39
@Krishna you may want to take a look at the API you're using (Apache?) and get a tutorial on how to parse the actualn response text, excluding headers. Clearly invoking `toString` won't work there. Overriding `toString` is likely not an option for you as this is not code you're owning. — Mena, Oct 16 '15 at 13:55

score 3 · Answer 1 · answered Oct 16 '15 at 13:27

3

Fairly certain response.toString doesn't do what you think it does, as it's not listed in the documentation.

I believe you need to use response.getEntity, and then entity.getContent, which gives you an InputStream to read the content from. Then pass that stream into your parser.

answered Oct 16 '15 at 13:27

T.J. Crowder

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Edited the question with modified code. Still same error using that code. Thanks for the immediate response. But can you help me further. – Krishna Oct 16 '15 at 13:37

score 1 · Accepted Answer · answered Oct 16 '15 at 14:58

Try this code. Use IOUtils as mentioned. it will work.

public class Check {
public static void main(String[] args){
String url = "http://localhost:8181/math/webapi/values";
HttpClient httpClient = HttpClientBuilder.create().build();
HttpGet request = new HttpGet(url);
try {
    HttpResponse response = httpClient.execute(request);
    InputStream value = response.getEntity().getContent();
    String jsonValue = IOUtils.toString(value);
    JSONObject json = new JSONObject(jsonValue);
    int i = json.getInt("x");
    System.out.println(i);
}catch (Exception e) {
    e.printStackTrace();
}       
}

org.json.JSONException: A JSONObject text must begin with '{' at character 1

2 Answers2