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Editorial Team
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Asked: 2026-06-16T17:28:54+00:00

I’m trying to send JSON data from a php script to an Android app

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I’m trying to send JSON data from a php script to an Android app and the php script’s output is different from what the Java app expects.

$data['sample']['txt']="hello world";
echo json_encode($data) // {"sample":{"txt":"hello world"}}
//above is incorrect, need {sample : [{txt:"hello world"}]}

The incorrect format results in the following Java exception: 

org.json.JSONException: Value {"txt":"hello world"} at sample of type org.json.JSONObject cannot be converted to JSONArray.

Is there an argument of PHP’s json_encode that I’m missing, or an alternative that would encode it properly?

Java code for async task:

public class RetrieveData extends AsyncTask<List<? extends NameValuePair>, Integer, List<String>> {

    protected List<String> doInBackground(List<? extends NameValuePair>... postData) {
        InputStream is = null;
        List<String> result = new ArrayList<String>();

        try {
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("http://192.168.72.2:10088/droid_test/test.php");
            httppost.setEntity(new UrlEncodedFormEntity(postData[0]));
            HttpResponse response = httpclient.execute(httppost);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();

            BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
            }
            reader.close();
            is.close();

            JSONObject JSONobj=new JSONObject(sb.toString());
            JSONArray JSONarr=JSONobj.getJSONArray("sample");

            for(int i = 0 ; i < JSONarr.length() ; i++){
                result.add(JSONarr.getJSONObject(i).getString("txt"));
            }

            }
        catch(Exception e) {
            result.add("ERROR "+e.toString());
        }

        return result;
    }

    protected void onPostExecute(List<String> result) {
            getHTTP(result);
    }

}

getHTTP(result) just sets the value to a TextView, which is where the error is displaying. (or the response if I hard code the echo statement)

Solution:

JSONObject JSONobj=new JSONObject(sb.toString());
JSONObject JSONarr=JSONobj.getJSONObject("sample"); // made object per @digitaljoel's suggestion

for(int i=0; i<JSONarr.length(); i++) {
    result.add(JSONarr.getString("txt")); // getting a String, not another array/object *duh*
}
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1 Answer

  1. Editorial Team

    Both of the JSON examples you show are valid JSON, it’s just a matter of what your mapping is on each end.

    Your java code is expecting “sample” to contain a collection (list or array) of objects where each object has a txt field. That’s why it has the [] around the object value in the JSON.

    You can change the mapping on the java side to expect only a single value for sample, or you can change the php code so $data[‘sample’] is an array with a single element that has ‘txt’ = “hello world”.

    If you include the mapping on the java side I can help with that. I’m sure some php guru could help if you want to fix it on the php side.

    Edit:

    JSONArray JSONarr=JSONobj.getJSONArray("sample"); is asking for an array. Change that to a JSONObject and you should be good to go.

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