conversion from string to JSON object Android

Question

I am working on an Android application. In my app I have to convert a string to JSON Object, then parse the values. I checked for a solution in Stackoverflow and found similar issue here link

The solution is like this

       `{"phonetype":"N95","cat":"WP"}`
        JSONObject jsonObj = new JSONObject("{\"phonetype\":\"N95\",\"cat\":\"WP\"}");

I use the same way in my code . My string is

{"ApiInfo":{"description":"userDetails","status":"success"},"userDetails":{"Name":"somename","userName":"value"},"pendingPushDetails":[]}

string mystring= mystring.replace("\"", "\\\"");

And after replace I got the result as this

{\"ApiInfo\":{\"description\":\"userDetails\",\"status\":\"success\"},\"userDetails\":{\"Name\":\"Sarath Babu\",\"userName\":\"sarath.babu.sarath babu\",\"Token\":\"ZIhvXsZlKCNL6Xj9OPIOOz3FlGta9g\",\"userId\":\"118\"},\"pendingPushDetails\":[]}

when I execute JSONObject jsonObj = new JSONObject(mybizData);

I am getting the below JSON exception

org.json.JSONException: Expected literal value at character 1 of

Please help me to solve my issue.

I guess the offending character is a backslash because of your substitution. Why exactly are you doing that? Where does the JSON string come from? — tiguchi
– tiguchi, Commented Aug 12, 2013 at 17:24
Just remove mystring= mystring.replace("\"", "\\\""); and see if it works for you then. — tiguchi
– tiguchi, Commented Aug 12, 2013 at 18:13

firelynx · Accepted Answer · 2015-08-14 07:34:40Z

264

Remove the slashes:

String json = {"phonetype":"N95","cat":"WP"};

try {

    JSONObject obj = new JSONObject(json);

    Log.d("My App", obj.toString());

} catch (Throwable t) {
    Log.e("My App", "Could not parse malformed JSON: \"" + json + "\"");
}

edited Aug 14, 2015 at 7:34

firelynx

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answered Aug 12, 2013 at 17:24

Phil

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10 Comments

Francisco Corrales Morales Over a year ago

what if the string is an array of JSON objects? Like "[{},{},{}]"

Phil Over a year ago

@FranciscoCorralesMorales you can use JSONArray obj = new JSONArray(json);. Then you can use a for-loop to iterate through the array.

Phil Over a year ago

@FranciscoCorralesMorales just use a try-catch block. If one fails, assume the other.

Phil Over a year ago

@ripDaddy69 It sounds like that is invalid JSON. It expects key-value pairings surrounded by curly brackets. Try something like {"Fat cat":"meow"}.

user1174868 Over a year ago

@Phil That doesn't appear to be a valid java String assignment. I don't understand what I am doing differently though JSONObject obj = new JSONObject("Fat cat":"meow"); I figured it out, I needed to use \ infront of the quotes, and then actual quotes around the whole thing. Thanks.

|

Serkan Hekimoglu · Accepted Answer · 2022-02-18 11:50:10Z

43

This method works

    String json = "{\"phonetype\":\"N95\",\"cat\":\"WP\"}";

    try {

        JSONObject obj = new JSONObject(json);

        Log.d("My App", obj.toString());
        Log.d("phonetype value ", obj.getString("phonetype"));

    } catch (Throwable tx) {
        Log.e("My App", "Could not parse malformed JSON: \"" + json + "\"");
    }

edited Feb 18, 2022 at 11:50

Serkan Hekimoglu

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answered May 14, 2016 at 20:39

Ercan ILIK

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2 Comments

CerebralFart Over a year ago

While this does answer the question, it doesn't explain why or how it works. Please add such an explanation.

kelalaka Over a year ago

It seems a simple code solution, that requires creating another object that handles the escape sequence.

Saro Taşciyan · Accepted Answer · 2014-03-13 10:42:49Z

8

try this:

String json = "{'phonetype':'N95','cat':'WP'}";

edited Mar 13, 2014 at 10:42

Saro Taşciyan

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answered Mar 13, 2014 at 10:22

user3139217

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3 Comments

Francisco Corrales Morales Over a year ago

what if the string is an array of JSON objects? Like "[{},{},{}]"

david m lee Over a year ago

This is a good idea. The single quote works and it eliminates the need for the escape characters.

Jesse Chisholm Over a year ago

Apostrophe might work, in JAVA, but it isn't strict legal JSON. So you may need to do things differently in other languages or situations.

Benson Githinji · Accepted Answer · 2020-09-10 08:12:28Z

You just need the lines of code as below:

   try {
        String myjsonString = "{\"phonetype\":\"N95\",\"cat\":\"WP\"}";
        JSONObject jsonObject = new JSONObject(myjsonString );
        //displaying the JSONObject as a String
        Log.d("JSONObject = ", jsonObject.toString());
        //getting specific key values
        Log.d("phonetype = ", jsonObject.getString("phonetype"));
        Log.d("cat = ", jsonObject.getString("cat");
    }catch (Exception ex) {
         StringWriter stringWriter = new StringWriter();
         ex.printStackTrace(new PrintWriter(stringWriter));
         Log.e("exception ::: ", stringwriter.toString());
    }

mhKarami · Accepted Answer · 2019-01-21 14:08:48Z

6

just try this , finally this works for me :

//delete backslashes ( \ ) :
            data = data.replaceAll("[\\\\]{1}[\"]{1}","\"");
//delete first and last double quotation ( " ) :
            data = data.substring(data.indexOf("{"),data.lastIndexOf("}")+1);
            JSONObject json = new JSONObject(data);

answered Jan 21, 2019 at 14:08

mhKarami

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Comments

RiccardoCh · Accepted Answer · 2016-11-10 08:10:26Z

To get a JSONObject or JSONArray from a String I've created this class:

public static class JSON {

     public Object obj = null;
     public boolean isJsonArray = false;

     JSON(Object obj, boolean isJsonArray){
         this.obj = obj;
         this.isJsonArray = isJsonArray;
     }
}

Here to get the JSON:

public static JSON fromStringToJSON(String jsonString){

    boolean isJsonArray = false;
    Object obj = null;

    try {
        JSONArray jsonArray = new JSONArray(jsonString);
        Log.d("JSON", jsonArray.toString());
        obj = jsonArray;
        isJsonArray = true;
    }
    catch (Throwable t) {
        Log.e("JSON", "Malformed JSON: \"" + jsonString + "\"");
    }

    if (object == null) {
        try {
            JSONObject jsonObject = new JSONObject(jsonString);
            Log.d("JSON", jsonObject.toString());
            obj = jsonObject;
            isJsonArray = false;
        } catch (Throwable t) {
            Log.e("JSON", "Malformed JSON: \"" + jsonString + "\"");
        }
    }

    return new JSON(obj, isJsonArray);
}

Example:

JSON json = fromStringToJSON("{\"message\":\"ciao\"}");
if (json.obj != null) {

    // If the String is a JSON array
    if (json.isJsonArray) {
        JSONArray jsonArray = (JSONArray) json.obj;
    }
    // If it's a JSON object
    else {
        JSONObject jsonObject = (JSONObject) json.obj;
    }
}

You could just test the first character of the JSON string to see if it is [ or { to know whether it is an array or an object. Then you wouldn't be risking both exceptions, just the pertinent one.
@JesseChisholm if you're sure that the string is a well formatted json array you can do what you say, otherwise you could risk and exception.

Thiago · Accepted Answer · 2021-02-20 00:14:10Z

2

Using Kotlin

    val data = "{\"ApiInfo\":{\"description\":\"userDetails\",\"status\":\"success\"},\"userDetails\":{\"Name\":\"somename\",\"userName\":\"value\"},\"pendingPushDetails\":[]}\n"
    
try {
      val jsonObject = JSONObject(data)
      val infoObj = jsonObject.getJSONObject("ApiInfo")
    } catch (e: Exception) {
    }

answered Feb 20, 2021 at 0:14

Thiago

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Comments

android.dk · Accepted Answer · 2015-02-09 13:11:56Z

Here is the code, and you can decide which
(synchronized)StringBuffer or faster StringBuilder to use.

Benchmark shows StringBuilder is Faster.

public class Main {
            int times = 777;
            long t;

            {
                StringBuffer sb = new StringBuffer();
                t = System.currentTimeMillis();
                for (int i = times; i --> 0 ;) {
                    sb.append("");
                    getJSONFromStringBuffer(String stringJSON);
                }
                System.out.println(System.currentTimeMillis() - t);
            }

            {
                StringBuilder sb = new StringBuilder();
                t = System.currentTimeMillis();
                for (int i = times; i --> 0 ;) {
                     getJSONFromStringBUilder(String stringJSON);
                    sb.append("");
                }
                System.out.println(System.currentTimeMillis() - t);
            }
            private String getJSONFromStringBUilder(String stringJSONArray) throws JSONException {
                return new StringBuffer(
                       new JSONArray(stringJSONArray).getJSONObject(0).getString("phonetype"))
                           .append(" ")
                           .append(
                       new JSONArray(employeeID).getJSONObject(0).getString("cat"))
                      .toString();
            }
            private String getJSONFromStringBuffer(String stringJSONArray) throws JSONException {
                return new StringBuffer(
                       new JSONArray(stringJSONArray).getJSONObject(0).getString("phonetype"))
                           .append(" ")
                           .append(
                       new JSONArray(employeeID).getJSONObject(0).getString("cat"))
                      .toString();
            }
        }

shaojun lyu · Accepted Answer · 2017-12-14 06:37:04Z

0

May be below is better.

JSONObject jsonObject=null;
    try {
        jsonObject=new JSONObject();
        jsonObject.put("phonetype","N95");
        jsonObject.put("cat","wp");
        String jsonStr=jsonObject.toString();
    } catch (JSONException e) {
        e.printStackTrace();
    }

answered Dec 14, 2017 at 6:37

shaojun lyu

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