Java parse JSON exception

Question

I am trying to parse a JSON String using Java.

This is my code so far:

    JSONObject obj = new JSONObject(Connection.getDataWOProxy(proxyUseQ));
    JSONArray arr = obj.getJSONArray("result");
    for (int i = 0; i < arr.length(); i++) {
        System.out.println(arr.getInt(i));
    }

Info: Connection.getDataWOProxy(proxyUseQ); is a method I wrote that gets the JSON String from an URL using a proxy. proxyUseQ is a boolean that tells the method to use the proxy or not.

What should my code do normally:

JSONObject obj = new JSONObject(Connection.getDataWOProxy(proxyUseQ));

^ This line initializes my JSONObject giving it the String. Works great.

JSONArray arr = obj.getJSONArray("result");

^ Here I want it to access the "result" tree - works as well if I get it right. Now, if I understood it correctly, all my information should be saved in that Array, is this correct?

    for (int i = 0; i < arr.length(); i++) {
        System.out.println(arr.getInt(i));
    }

^ Now, I want to access the several information and print e.g. the short_description from result 2 (Console output should be: TEST 2). This is where my code fails:

org.json.JSONException: JSONArray[0] is not a number.

My JSON String:

{
  "result": [
    {
      "number": "1234",
      "short_description": "TEST",
      "priority": "4 - Low",
      "caller_id": "Some, User"
    },
    {
      "number": "12345",
      "short_description": "TEST 2",
      "priority": "4 - Low",
      "caller_id": "Some, User2"
    }
  ]
}

What am I doing wrong? I am new to JSON. Thanks.

shmosel · Accepted Answer · 2017-03-15 11:51:03Z

1

It's an array of JSON objects; they can't be converted to int. If you want to access a specific field on the current object, you can do so like this:

System.out.println(arr.getJSONObject(i).getString("short_description"));

answered Mar 15, 2017 at 11:51

shmosel

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Comments

Arpan Sharma · Accepted Answer · 2017-03-15 11:52:32Z

1

Yes because its not string or number its whole

{
      "number": "1234",
      "short_description": "TEST",
      "priority": "4 - Low",
      "caller_id": "Some, User"
    },

to get the number you have to

arr.getJSONObject(i).getString("number");

answered Mar 15, 2017 at 11:52

Arpan Sharma

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Comments

S. Stas · Accepted Answer · 2017-03-15 12:05:32Z

0

JSONArray[0] is not a number because it is an object. Based on your JSON String, you have array named "result", that contains two elements. arr.getInt(i) return the JSONArray[0] as JSONObject. You should call JSONObject obj = JSONArray.getJSONObject(i) and then you should call obj.getInt(i) Note, that your json fields are generally strings, not integers.

answered Mar 15, 2017 at 12:05

S. Stas

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Comments

Rajesh Gopu · Accepted Answer · 2017-03-15 12:09:10Z

arr.getInt(i) is wrong ,its first entry of "result" key of type JSONObject ,so getInt on JSONObject is invalid .

Below is the right procedure , to get only description

JSONArray arr = obj.getJSONArray("result");
for (int i = 0; i < arr.length(); i++) {
    JSONObject entry = arr.getJSONObject(i);
    System.out.println(entry.getString("short_description"));
}

ouput:

TEST
TEST 2

to print each key & values.

    JSONArray arr = obj.getJSONArray("result");
    for (int i = 0; i < arr.length(); i++) {
        JSONObject entry = arr.getJSONObject(i);
        Iterator<String> keyIter= entry.keys();
        System.out.println("---------------->Entry :" + (i+1));
        while(keyIter.hasNext()){
            String key = keyIter.next();
            System.out.print(key + " ==> ");
            System.out.print(entry.get(key)+"\n");
        }
    }

Output is

---------------->Entry :1
number ==> 1234
short_description ==> TEST
caller_id ==> Some, User
priority ==> 4 - Low
---------------->Entry :2
number ==> 12345
short_description ==> TEST 2
caller_id ==> Some, User2
priority ==> 4 - Low

Mateo · Accepted Answer · 2019-10-04 19:23:13Z

0

 //import java.util.ArrayList;
 //import org.bson.Document;


 Document root = Document.parse("{ \"result\" : [{ \"number\" : \"1234\", \"short_description\" : \"TEST\", \"priority\" : \"4 - Low\", \"caller_id\" : \"Some, User\" }, { \"number\" : \"12345\", \"short_description\" : \"TEST 2\", \"priority\" : \"4 - Low\", \"caller_id\" : \"Some, User2\" }] }");

 System.out.println(((String)((Document)((ArrayList)root.get("result")).get(0)).get("number")));
 System.out.println(((String)((Document)((ArrayList)root.get("result")).get(0)).get("short_description")));
 System.out.println(((String)((Document)((ArrayList)root.get("result")).get(0)).get("priority")));
 System.out.println(((String)((Document)((ArrayList)root.get("result")).get(0)).get("caller_id")));
 System.out.println(((String)((Document)((ArrayList)root.get("result")).get(1)).get("number")));
 System.out.println(((String)((Document)((ArrayList)root.get("result")).get(1)).get("short_description")));
 System.out.println(((String)((Document)((ArrayList)root.get("result")).get(1)).get("priority")));
 System.out.println(((String)((Document)((ArrayList)root.get("result")).get(1)).get("caller_id")));

answered Oct 4, 2019 at 19:23

Mateo

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2 Comments

OrdoFlammae Over a year ago

Could you add an explanation of the problem you code fixes from the original?

Mateo Over a year ago

This is an example of how to do it with the full JSON as apposed to just the snippet. It also shows how to do it with BSON if you are working with MongoDB.

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