Question

我收到此错误

Error in parsing Dataorg.json.JSONException: Value ["AACHI"] at 0 of type org.json.JSONArray cannot be converted to JSONObject

Process: info.androidhive.materialdesign, PID: 18373
java.lang.NullPointerException: Attempt to invoke virtual method 'java.lang.Object android.content.Context.getSystemService(java.lang.String)' on a null object reference

这是我试图解析的json数据。[[＆＃34; AACHI＆＃34;]，[＆＃34; AJAY COMPANY＆＃34;]，[＆＃34; ALL OUT＆＃34; ]，[＆＃34; AMBICA＆＃34;]，[＆＃34; AMICO PRODUCTS＆＃34;]，[＆＃34; AMUL＆＃34;]，[＆＃34; ANAND BHOG PRODUCTS＆＃34;]，[ ＆＃34; ANNAPURNA＆＃34;]，[＆＃34; ANOOS PRODUCTS＆＃34;]，[＆＃34; ARVIND LABORATORIES＆＃34;]，[＆＃34; ASWINI PHARMACEUTICALS＆＃34;]，[＆＃ 34; ATTK PRODUCTS＆＃34;]，[＆＃34; AVA＆＃34;]，[＆＃34; BAJAJ PRODUCTS＆＃34;]，[＆＃34; BAMBINO＆＃34;]，[＆＃34; BANJARAS产品＆＃34;]，[＆＃34; BHAGAYALAKSHIMI PRODUCTS＆＃34;]，[＆＃34; BRITANNIA＆＃34;]，[＆＃34; cadbury＆＃34;]]

这是我解析json数据的代码片段。

 class DownloadJson extends AsyncTask {

    Activity context;
    ListView myListView;

    private  ProgressDialog dialog = new ProgressDialog(getActivity());
    public DownloadJson(Activity context, ListView myListView) {
        this.myListView = myListView;
        this.context = context;

    }

    public DownloadJson() {

    }


    @Override
    protected void onPreExecute() {

        this.dialog.setMessage("Please wait");
        this.dialog.show();
    }

    @Override
    protected void onCancelled() {
        super.onCancelled();
    }

    @Override
    protected Object doInBackground(Object[] params) {
        String result = null;
        InputStream isr = null;
        String imageId = null;
        String ip = "http://ganarajsshetti.tk/mobileapp/selectjson.php/";
        try {

            HttpClient httpClient = new DefaultHttpClient();
            HttpGet httpGet = new HttpGet(ip);
            HttpResponse httpResponse = httpClient.execute(httpGet);
            HttpEntity httpEntity = httpResponse.getEntity();
            isr = httpEntity.getContent();
        } catch (Exception e) {
            Log.e("log_tag", "Error in http Connection" + e.toString());

        }
        // converting response to string
        try {
            BufferedReader br = new BufferedReader(new InputStreamReader(isr));
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = br.readLine()) != null) {
                sb.append(line + "\n");
            }
            isr.close();
            result = sb.toString();

        } catch (Exception e) {
            Log.e("log_tag", "Error in Converting Data" + e.toString());
        }

        // parse JSON data

        try {

            JSONArray jsonArray = new JSONArray(result);
            strListView = new String[jsonArray.length()];
            for (int i = 0; i < jsonArray.length(); i++) {
               JSONObject json_data = jsonArray.getJSONObject(i);

                strListView[i] = json_data.getString("Num_Rows");

                System.out.println("--------------"+json_data.getString("Num_Rows"));

                Log.e("ACK_tag", "DATA" + strListView[i]);
            }

        } catch (Exception e) {
            Log.e("log_tag", "Error in parsing Data" + e.toString());
        }

        return null;

    }

    @Override
    protected void onPostExecute(Object o) {
        //objAdapter = new ArrayAdapter<String>(context,R.layout.vendorwithimage,R.id.venderdescrip,strListView);
        if (dialog.isShowing()) {
            dialog.dismiss();
        }
        callCustomAdaper(context);


    }
}
@Override
public void onStop() {
    super.onStop();
    if (task.getStatus() == AsyncTask.Status.RUNNING)
        task.cancel(true);
    if (task.getStatus() == AsyncTask.Status.RUNNING)
        task.cancel(true);

}

`

Answer 1

它的json数组数组。它就像一个2D阵列。

JsonArray jsonarray= new JsonArray(json);
 for (int i = 0; i < jsonArray.length(); i++) {
  strListView[i] = jsonarray.getJsonArray(i);


}

链接来源：https://www.json.com/json-array

Answer 2

JSONObject json_data = jsonArray.getJSONObject（i）;

这里＆＃39; jsonArray＆＃39;包含数组对象而不是json对象。所以你必须把它改成JsonArray

  JSONArray jsonArrayElement = jsonArray.getJSONArray(0);

此json也有效，但不是常规或有用.. !!

查看json here的更多信息。

Answer 3

尝试更改行

JSONObject json_data = jsonArray.getJSONObject(i);
strListView[i] = json_data.getString("Num_Rows");

到

JSONArray innerJsonArray = jsonArray.getJSONArray(i);
strListView[0] = json_data.get(0);

Answer 4

试试这个。工作正常

String json = "[[\"AACHI\"],[\"AJAY COMPANY\"],[\"ALL OUT\"],[\"AMBICA\"],[\"AMICO PRODUCTS\"],[\"AMUL\"],[\"ANAND BHOG PRODUCTS\"],[\"ANNAPURNA\"],[\"ANOOS PRODUCTS\"],[\"ARVIND LABORATORIES \"],[\"ASWINI PHARMACEUTICALS\"],[\"ATTK PRODUCTS\"],[\"AVA\"],[\"BAJAJ PRODUCTS \"],[\"BAMBINO\"],[\"BANJARAS PRODUCTS\"],[\"BHAGAYALAKSHIMI PRODUCTS\"],[\"BRITANNIA\"],[\"cadbury\"]]";
     try {
        JSONArray jsAr = new JSONArray(json);
        String job="";
        for(int i=0;i<jsAr.length();i++)
        {
            JSONArray js1 = jsAr.getJSONArray(i);
            job += js1.getString(0);

        }
        Toast.makeText(this,job,Toast.LENGTH_LONG).show();

    } catch (JSONException e) {
        // TODO Auto-generated catch block

        Toast.makeText(this,"Error",Toast.LENGTH_LONG).show();
        e.printStackTrace();
    }

在将JSON数据解析为android时获取错误

4 个答案: