机器人 - 获得来自服务器的响应时,为了避免一个JSONException验证JSON [英] Android - Validating JSON when getting a response from a server to avoid a JSONException

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问题描述

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<太虚,太虚,字符串>
{
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<&的NameValuePair GT; namevaluepairs中=新的ArrayList<&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< 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<一个链路href=\"http://stackoverflow.com/questions/10587599/how-do-i-$p$pvent-my-app-from-crashing-unexpectedly-force-close-when-using-js?rq=1\">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)

解决方案

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()); 
    }   
}

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