php - JSON 异常没有返回结果

Question

Android 应用程序从 PHP 文件中接收 ID、名称和电子邮件 ID、名称、电子邮件作为 JSON 响应发送到 Android 应用程序，问题是当我单击“获取数据”时没有收到数据并且我在 LogCat 中收到异常 这是我的代码、PHP 文件和堆栈跟踪 主要 Activity

   package com.example.colcheck;

import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;
import android.app.Activity;
import android.app.ProgressDialog;
import android.os.AsyncTask;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.TextView;

import com.example.colcheck.Library.*;

public class MainActivity extends Activity {
    TextView uid;
    TextView name1;
    TextView email1;
    Button Btngetdata;

    //URL to get JSON Array
    private static String url = "http://localhost/new/check2.php";

    //JSON Node Names 
    private static final String TAG_USER = "user";
    private static final String TAG_ID = "id";
    private static final String TAG_NAME = "name";
    private static final String TAG_EMAIL = "email";

    JSONArray user = null;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);

        setContentView(R.layout.activity_main);
        Btngetdata = (Button)findViewById(R.id.getdata);
        Btngetdata.setOnClickListener(new View.OnClickListener() {

            @Override
            public void onClick(View view) {
                 new JSONParse().execute();

            }
        });

    }

    private class JSONParse extends AsyncTask<String, String, JSONObject> {
         private ProgressDialog pDialog;
        @Override
        protected void onPreExecute() {
            super.onPreExecute();
             uid = (TextView)findViewById(R.id.uid);
             name1 = (TextView)findViewById(R.id.name);
             email1 = (TextView)findViewById(R.id.email);
            pDialog = new ProgressDialog(MainActivity.this);
            pDialog.setMessage("Getting Data ...");
            pDialog.setIndeterminate(false);
            pDialog.setCancelable(true);
            pDialog.show();

        }

        @Override
        protected JSONObject doInBackground(String... args) {
            JSONParser jParser = new JSONParser();

            // Getting JSON from URL
            JSONObject json = jParser.getJSONFromUrl(url);
            return json;
        }
         @Override
         protected void onPostExecute(JSONObject json) {
             pDialog.dismiss();
             try {

                 System.out.println("JSON Response : "+json.toString().trim());

                 String id="";
                 String name="";
                 String email="";
                if(json.has("id"))
                     id = json.getString("id");

                 if(json.has("name"))
                      name = json.getString("name");

                 if(json.has("TAG_EMAIL"))
                      email = json.getString("TAG_EMAIL");

                                 //Set JSON Data in TextView
                                 uid.setText(id);
                                 name1.setText(name);
                                 email1.setText(email);

                         } catch (JSONException e) {
                             e.printStackTrace();
                         }

         }
    }

}

带有 json 响应的 PHP 文件

 <?php

    $con=mysqli_connect("localhost","root","123","pet_home");
    if (mysqli_connect_errno())
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }

   $result = mysqli_query($con,"SELECT * FROM users WHERE username='test' AND password='123'");
    $row_cnt = mysqli_num_rows($result);

     if($row_cnt>0){     
         $row = mysqli_fetch_array($result); 
         $data = array('success'=>true, 'user'=>array("TAG_ID" => $row['id'], "TAG_NAME"=> $row['name'], "TAG_EMAIL" =>$row['email'])); 
     }else{
         $data = array('success'=>false);
     }

header('Content-Type: application/json');
   echo json_encode($data);



    mysqli_close($con);


    ?>

执行此操作后没有响应返回到 android 应用程序这里是堆栈跟踪似乎有异常

11-30 17:06:13.679: I/System.out(3474): JSON Response : {"TAG_USER":{"id":"OFF","TAG_EMAIL":"OFF","name":"OFF"},"success":true}

Answer 1

试试这个..

如果你的json是这样的

{"id":"OFF","TAG_EMAIL":"OFF","name":"OFF"}

然后

您可以检查该标签名称是否存在

System.out.println("JSON Response : "+json.toString().trim());

if(json.has("id"))
    String id = json.getString("id");

if(json.has("name"))
    String name = json.getString("name");

if(json.has("TAG_EMAIL"))
    String email = json.getString("TAG_EMAIL");

编辑:

你的解析如下

user = json.getJSONArray(TAG_USER);

没有 JSONArray 只有 JSONObject 所以你不能将 JSONObject 解析为 JSONArray 然后没有 用户作为JSONArray

编辑 1:

try {

System.out.println("JSON Response : "+json.toString().trim());

if(json.has("id"))
    String id = json.getString("id");

if(json.has("name"))
    String name = json.getString("name");

if(json.has("TAG_EMAIL"))
    String email = json.getString("TAG_EMAIL");

                //Set JSON Data in TextView
                uid.setText(id);
                name1.setText(name);
                email1.setText(email);

        } catch (JSONException e) {
            e.printStackTrace();
        }

编辑 2:

try {

System.out.println("JSON Response : "+json.toString().trim());

 JSONObject c = json.getJSONObject("TAG_USER");

if(c.has("id"))
    String id = c.getString("id");

if(c.has("name"))
    String name = c.getString("name");

if(c.has("TAG_EMAIL"))
    String email = c.getString("TAG_EMAIL");

                //Set JSON Data in TextView
                uid.setText(id);
                name1.setText(name);
                email1.setText(email);

        } catch (JSONException e) {
            e.printStackTrace();
        }