java.lang.String cannot be cast to org.json.simple.JSONObject simple-json

Question

I am getting strange problem while trying to parse a simple json using simple-json by google.

Here is my code which is not working:

String s = args[0].toString();
JSONObject json = (JSONObject)new JSONParser().parse(s);

When I execute, it will give me the exception java.lang.String cannot be cast to org.json.simple.JSONObject

But when I hard code json directly like below its working fine. Wat could be the reason?

JSONObject json = (JSONObject)new JSONParser().parse("{\"application\":\"admin\",\"keytype\":\"PRODUCTION\",\"callbackUrl\":\"qwerewqr;ewqrwerq;qwerqwerq\",\"authorizedDomains\":\"ALL\",\"validityTime\":\"3600000\",\"retryAfterFailure\":true}");

UPDATE

When I print s, it will give me the output below:

"{\"application\":\"admin\",\"keytype\":\"PRODUCTION\",\"callbackUrl\":\"qwerewqr;ewqrwerq;qwerqwerq\",\"authorizedDomains\":\"ALL\",\"validityTime\":\"3600000\",\"retryAfterFailure\":true}"

Maybe it's because s was in a wrong format? Or do you get a different error in that case? In your second example, try to put in an invalid JSON string, like "test123" and see the error. — Philipp Murry
– Philipp Murry, Commented Aug 29, 2014 at 11:08
Could you change and debug it with Object obj = new JSONParser().parse(s) instead of JSONObject json = (JSONObject)new JSONParser().parse(s) — Devrim
– Devrim, Commented Aug 29, 2014 at 11:29
@aegean yes that is wroking.. thanks i just wondering how could I get print the json elements with it? — Roshan Wijesena
– Roshan Wijesena, Commented Aug 29, 2014 at 11:36
@PhilippMurry when i am giving "test123" its giving Unexpected character (t) at position 0. — Roshan Wijesena
– Roshan Wijesena, Commented Aug 29, 2014 at 11:48

mooncat39 · Accepted Answer · 2015-08-15 14:41:35Z

7

I ran this through eclipse by providing arguments in run configuration.

 public static void main(String[] args) {
        String s = args[0].toString();
        System.out.println("=>" + s);
        try {
            JSONObject json = (JSONObject) new JSONParser().parse(s);
            System.out.println(json);
        } catch (ParseException e) {
            e.printStackTrace();
        }
    }

Output

=>{"application":"admin","keytype":"PRODUCTION","callbackUrl":"qwerewqr;ewqrwerq;qwerqwerq","authorizedDomains":"ALL","validityTime":"3600000","retryAfterFailure":true}


{"validityTime":"3600000","callbackUrl":"qwerewqr;ewqrwerq;qwerqwerq","application":"admin","retryAfterFailure":true,"authorizedDomains":"ALL","keytype":"PRODUCTION"}

edited Aug 15, 2015 at 14:41

mooncat39

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answered Aug 29, 2014 at 11:23

Ankur Singhal

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6 Comments

Roshan Wijesena Over a year ago

thanks .. this is working but my problem is when i invoke from args[0].toString(); then not working..

Hot Licks Over a year ago

@RoshanWijesena - Print out what you get from args[0]. You SHOULD NOT see any escapes. Presumably parse returns a String error message (which you should test with instanceof) when it encounters an error.

Roshan Wijesena Over a year ago

@HotLicks thanks. When i am print it out it is giving me "{\"application\":\"admin\",\"keytype\":\"PRODUCTION\",\"callbackUrl\":\"qwerewqr;ewqrwerq;qwerqwerq\",\"authorizedDomains\":\"ALL\",\"validityTime\":\"3600000\",\"retryAfterFailure\":true}"

Ankur Singhal Over a year ago

@RoshanWijesena answer edited, as per your requirement

Hot Licks Over a year ago

Yep, the escapes do not belong there. It isn't a C string, so escapes may not be required (depending on the OS and command analyzer used).

|

TanvirChowdhury · Accepted Answer · 2021-10-05 07:29:03Z

2

Make sure that the string is a valid JSON. You can user JSONObject parameterized constructor with the given string to convert the JSON string to a valid JSON object.

For example,

try {
    String jsonString = " {'application':'admin','keytype':'PRODUCTION','callbackUrl':'qwerewqr;ewqrwerq;qwerqwerq','authorizedDomains':'ALL','validityTime':3600000,'retryAfterFailure':true}";
 
    JSONObject data = new JSONObject(jsonString);
    String application = data.getString("application"); //gives admin
    String keytype = data.getString("keytype"); //gives PRODUCTION
} catch (JSONException e) {
    e.printStackTrace();
}

edited Oct 5, 2021 at 7:29

TanvirChowdhury

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answered Jan 28, 2021 at 8:33

Codemaker2015

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Comments

Mohammad Noman · Accepted Answer · 2020-09-04 18:49:58Z

1

I had the same issue


   package com.test;

   import org.json.JSONObject;
   import org.json.simple.parser.JSONParser;
   import org.json.simple.parser.ParseException;

   public class JSONTest {    

        public static void main(String[] args) {

            String s = args[0];
            try {
                JSONObject json = new JSONObject((String) new JSONParser().parse(s));
                System.out.println(json);
            } catch (ParseException e) {
                e.printStackTrace();
            }
      }
   }

This worked for me

answered Sep 4, 2020 at 18:49

Mohammad Noman

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Comments

Shailendra · Accepted Answer · 2014-08-29 11:54:24Z

Try this

package com.test;

import org.json.simple.JSONObject;
import org.json.simple.parser.JSONParser;
import org.json.simple.parser.ParseException;

public class JSONTest {    

        public static void main(String[] args) {

            String s = args[0];
            try {
                JSONObject json = (JSONObject) new JSONParser().parse(s);
                System.out.println(json);
            } catch (ParseException e) {
                e.printStackTrace();
            }
        }

Then on command line

java -classpath  ".;json-simple-1.1.1.jar" com.test.JSONTest {\"application\":\"admin\",\"keytype\":\"PRODUCTION\",\"callbackUrl\":\"qwerewqr;ewqrwerq;qwerqwerq\",\"authorizedDomains\":\"ALL\",\"validityTime\":\"3600000\",\"retryAfterFailure\":true}

The out put is

{"validityTime":"3600000","callbackUrl":"qwerewqr;ewqrwerq;qwerqwerq","application":"admin","retryAfterFailure":true,"authorizedDomains":"ALL","keytype":"PRODUCTION"}

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