在我的应用程序中，我有一个食物活动，用户在其中输入食物，该应用程序通过用户输入的名称从MYSQL数据库中请求食物。如果输入的食物不存在，则数据库返回的字符串应为null。

当前，发生这种情况时，将发生异常，因为无法将null值解析为JSON数组。我的问题是：“有没有办法防止我的应用程序强制关闭？我可以处理该异常并显示一个吐司消息，通知用户未找到所请求的食物吗？”
我想防止应用程序崩溃，而可以正常地失败。

请帮我。

我已经在我的应用程序中显示了相关代码。

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()); 
        }   
    }
}