在我的一些应用程序，跟一个服务器，并使用http我使用JSON数据服务器端格式化，当它到达的设备我用类似这样的code，我的计算器发现了一些得到回应

 私有类LoadData扩展的AsyncTask＆LT;太虚，太虚，字符串＆GT;{私人JSONArray jArray;私人字符串结果= NULL;私人InputStream为= NULL;。私有String entered_food_name = choice.getText（）的toString（）修剪（）;在preExecute保护无效（）{}@覆盖保护字符串doInBackground（无效... PARAMS）{尝试{ArrayList的＆LT;＆的NameValuePair GT; namevaluepairs中=新的ArrayList＆LT;＆的NameValuePair GT;（）;HttpClient的HttpClient的=新DefaultHttpClient（）;HttpPost httppost =新HttpPost（http://10.0.2.2/food.php）;nameValuePairs.add（新BasicNameValuePair（名，entered_food_name））;httppost.setEntity（新UrlEn codedFormEntity（namevaluepairs中，UTF-8））;HTT presponse响应= httpclient.execute（httppost）;HttpEntity实体= response.getEntity（）; 是= entity.getContent（）;    //响应转换为字符串读者的BufferedReader =新的BufferedReader（新的InputStreamReader（是，UTF-8），8）;    StringBuilder的SB =新的StringBuilder（）;    串线= NULL;    而（（行= reader.readLine（））！= NULL）{        sb.append（线）;    }    is.close（）;    结果= sb.toString（）;    。结果= result.replace（'\\'，'\\''）修剪（）;}赶上（例外五）{    Log.e（log_tag，连接+ e.toString（））;}返回结果;}@覆盖保护无效onPostExecute（字符串结果）{尝试{    串foodName =;    int描述= 0;    jArray =新JSONArray（结果）; //此处若结果为空将发生exeption    JSONObject的json_data = NULL;    的for（int i = 0; I＆LT; jArray.length（）;我++）{        json_data = jArray.getJSONObject（ⅰ）;        foodName = json_data.getString（名称）;        。        。        。        。        。    }    赶上（JSONException E）{        ** //我可以做什么这里$ P $坠机pvent我的应用程序和        //使土司输入食品可用状态并没有???? **        Log.e（log_tag，parssing错误+ e.toString（））;    }}} 

我试图用在哪里，我发现这个code页面上的解决方案。继承人顺便How我prevent我的应用程序意外崩溃，＆QUOT;强制关闭＆QUOT ;,使用JSON数据时，并处理异常，而不是

我试图检查我的在线验证JSON响应并返回JSON是有效的。林想，当用户在一个贫穷的覆​​盖区域的连接立即中断或什么导致JSON被打破，导致即使我已经试过在我张贴在该链接的线程所建议的方法的JSON异常。所以，我希望能够验证JSON当它到达的电话，以便如果有效，我可以进行，如果不是我想尝试修复它，或者至少不会试图通过一个破碎和格式不正确的JSON阵列成引起该JSONException的方法之一。

在验证字符串我在得到响应，以确保其有效的JSON将是真棒任何提示。

编辑：有关问题的补充堆栈转储

  org.json.JSONException：在类型的结果值null org.json.JSONObject $ 1不能被转换成JSONArray在org.json.JSON.typeMismatch（JSON.java:96）在org.json.JSONObject.getJSONArray（JSONObject.java:548）在it.cores.Activity $ GetM.doInBackground（Activity.java:1159）在it.cores.Activity $ GetM.doInBackground（Activity.java:1）在android.os.AsyncTask $ 2.call（AsyncTask.java:185）在java.util.concurrent.FutureTask中$ Sync.innerRun（FutureTask.java:306）在java.util.concurrent.FutureTask.run（FutureTask.java:138）在java.util.concurrent.ThreadPoolExecutor.runWorker（ThreadPoolExecutor.java:1088）在java.util.concurrent.ThreadPoolExecutor中的$ Worker.run（ThreadPoolExecutor.java:581）在java.lang.Thread.run（Thread.java:1019） 

解决方案

您必须确保你做一个空检查/或任何其他验证如检查后做了解析，如果它包含响应code做任何之前在字符串结果操作。

 保护无效onPostExecute（字符串结果）{尝试{    串foodName =;    int描述= 0;如果（结果！= NULL）//做你进行结果验证{    jArray =新JSONArray（结果）; //此处若结果为空将发生exeption    JSONObject的json_data = NULL;    的for（int i = 0; I＆LT; jArray.length（）;我++）{        json_data = jArray.getJSONObject（ⅰ）;        foodName = json_data.getString（名称）;        。        。        。        。        。    } }   赶上（JSONException E）{        ** //我可以做什么这里$ P $坠机pvent我的应用程序和        //使土司输入食品可用状态并没有???? **        Log.e（log_tag，parssing错误+ e.toString（））;    }} 

In some of my applications that talk to a server and get a response using http I use json to format the data server side and when it gets to the device I use something similar to this code which I found on stackoverflow:

private class LoadData extends AsyncTask<Void, Void, String> 
{ 
private  JSONArray jArray;
private  String result = null;
private  InputStream is = null;
private String entered_food_name=choice.getText().toString().trim();
protected void onPreExecute() 
{
}

@Override
protected String doInBackground(Void... params) 
{
try {
ArrayList<NameValuePair> nameValuePairs = new            ArrayList<NameValuePair>();
HttpClient httpclient = new DefaultHttpClient(); 
HttpPost httppost = new HttpPost("http://10.0.2.2/food.php");
nameValuePairs.add(new BasicNameValuePair("Name",entered_food_name));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs,"UTF-8"));
HttpResponse response = httpclient.execute(httppost);

HttpEntity entity = response.getEntity();
 is = entity.getContent();

    //convert response to string

BufferedReader reader = new BufferedReader(new InputStreamReader(is,"utf-8"),8);
    StringBuilder sb = new StringBuilder();
    String line = null;
    while ((line = reader.readLine()) != null) {
        sb.append(line);
    }

    is.close();


    result =sb.toString();
    result = result.replace('\"', '\'').trim();

}
catch(Exception e){
    Log.e("log_tag", " connection" + e.toString());                     
}


return result;  

}

@Override
protected void onPostExecute(String result) 
{  
try{

    String foodName="";
    int Description=0;

    jArray = new JSONArray(result); // here if the result is null an exeption will occur
    JSONObject json_data = null;

    for (int i = 0; i < jArray.length(); i++) {
        json_data = jArray.getJSONObject(i);
        foodName=json_data.getString("Name");
        .
        .
        .
        .
        .
    } 
    catch(JSONException e){ 
        **// what i can do here to prevent my app from crash and 
        //  make toast " the entered food isnot available " ????**
        Log.e("log_tag", "parssing  error " + e.toString()); 
    }   
}
}

I tried to use the solution on the page where I found this code. heres the link by the way How do I prevent my app from crashing unexpectedly, "force close", when using JSON data, and handle the exception instead?

I tried checking my json response with online validators and the json returned is valid. Im thinking that when users are in a poor coverage area the connection momentarily breaks or something causes the json to be broken which causes a json exception even though I have tried the method suggested in the thread I posted in that link. So I would like to be able to validate the JSON when it gets to the phone so if its valid I can proceed and if not than I would like to try and fix it or at least not try to pass a broken and not properly formatted json array into one of the methods that causes the JSONException.

Any tips on validating the string I get in the response to make sure its valid JSON would be awesome.

Edit: added stack dump related to problem

org.json.JSONException: Value null at results of type org.json.JSONObject$1 cannot be     converted to JSONArray
at org.json.JSON.typeMismatch(JSON.java:96)
at org.json.JSONObject.getJSONArray(JSONObject.java:548)
at it.cores.Activity$GetM.doInBackground(Activity.java:1159)
at it.cores.Activity$GetM.doInBackground(Activity.java:1)
at android.os.AsyncTask$2.call(AsyncTask.java:185)
at java.util.concurrent.FutureTask$Sync.innerRun(FutureTask.java:306)
at java.util.concurrent.FutureTask.run(FutureTask.java:138)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1088)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:581)
at java.lang.Thread.run(Thread.java:1019)

You have to make sure you do the parsing after doing a null check/or any other validation like checking if it contains the response code before doing any operation on the String result.

protected void onPostExecute(String result) 
{  
try{

    String foodName="";
    int Description=0;

if(result!=null)//Do your validation for result

{
    jArray = new JSONArray(result); // here if the result is null an exeption will occur
    JSONObject json_data = null;

    for (int i = 0; i < jArray.length(); i++) {
        json_data = jArray.getJSONObject(i);
        foodName=json_data.getString("Name");
        .
        .
        .
        .
        .
    } 
 }
   catch(JSONException e){ 
        **// what i can do here to prevent my app from crash and 
        //  make toast " the entered food isnot available " ????**
        Log.e("log_tag", "parssing  error " + e.toString()); 
    }   
}