org.json.JSONException无法解析android中的json响应

Question

我正在开发一个需要用户登录名和密码的应用程序。我的PHP脚本运行完美。但是我的Android代码存在一些问题。在记录中我遇到了堆栈跟踪。这里是我的代码

public class MainActivity extends AppCompatActivity 
{ 
EditText userName; 
EditText password; 
Button sButton; 
HttpClient httpClient; 
HttpPost httpPost; 
HttpResponse httpResponse; 
String username; 
String pass; 
@Override 
protected void onCreate(Bundle savedInstanceState) 
{ 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.activity_main); 
    userName = (EditText)findViewById(R.id.user_id); 
    password = (EditText) findViewById(R.id.user_password); 
    sButton= (Button) findViewById(R.id.s_button); 
    username = userName.getText().toString(); 
    pass = password.getText().toString(); 
    httpClient = new DefaultHttpClient(); 

    httpPost = new HttpPost("http://192.168.100.106/EMS/functions.php"); 
    final JSONObject jsonObject = new JSONObject(); 


    sButton.setOnClickListener(new View.OnClickListener() { 
     @Override 
     public void onClick(View view) 
     { 
       Thread thread = new Thread() 
       { 
        @Override 
        public void run() 
        { 
         try 

         { 
          jsonObject.put("username", username); 
          jsonObject.put("password", pass); 
          Log.wtf("Sent data :","username and password"); 
          httpPost.setEntity(new StringEntity(jsonObject.toString())); 
         } 

         catch(JSONException | UnsupportedEncodingException e) 

         { 
          e.printStackTrace(); 
         } 
         try { 
          httpResponse = httpClient.execute(httpPost); 
          Log.wtf("Request sent: "," httpresponse"); 
          String str = EntityUtils.toString(httpResponse.getEntity()); 
          Log.wtf("String recieved ",str); 
          JSONObject responseObject = new JSONObject(str); 
          String response = responseObject.getString("success"); 
          Log.wtf("Response recieved ",response); 
          if(response.equals("1")) 
          { 
           runOnUiThread(new Runnable() { 
            @Override 
            public void run() { 
             Toast.makeText(getApplicationContext(), "Credentials match successful.",Toast.LENGTH_SHORT).show(); 
             Intent intent = new Intent(MainActivity.this,index.class); 
             startActivity(intent); 

            } 
           }); 
          } 

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

        } 
       };thread.start(); 



     } 
    }); 
} 

} 

这里是我的堆栈跟踪

08-08 23:31:46.726 17362-17472/milind.com.ems A/Sent data :﹕ username and password 
08-08 23:31:46.771 17362-17472/milind.com.ems A/Request sent:﹕ httpresponse 
08-08 23:31:46.811 17362-17472/milind.com.ems A/String recieved﹕ [ 08-08  23:31:46.812 17362:17472 W/System.err ] 
org.json.JSONException: End of input at character 2 of 

我的PHP脚本是：

<?php 

class functions 
{ 
private $con; 

function __construct() 
{ 
    require_once 'DB_Connect.php'; 
    $this->db = new DB_Connect(); 
    $this->con =$this->db->connect(); 
} 
function __destruct() 
{ 

} 

function getUser() 
{ 
    $json= file_get_contents("php://input"); 
    $str = iconv('UTF-8', 'UTF-8//IGNORE', utf8_encode($json)); 
    $str = json_decode($str,true); 
    $ID = $str['username']; 
    $password = $str['password']; 
    $password = $this->clearstring($password); 
    // echo "Password :- " . $password; 
    $q="SELECT * FROM employers WHERE Employers_ID = '{$ID}' ";   
    $user =$this->con->query($q);   
    //echo mysqli_error($this->con); 

    if($user->num_rows >0) 
    { 
     $row = $user->fetch_assoc();   
     $db_password = $row['Password'];    
     $this->compare($db_password,$password); 
    } 

    } 
    function clearstring($str) 
    { 
     //$str = strip_tags($str); 
     $str = stripcslashes($str); 
     $str = htmlspecialchars($str); 
     $str = trim($str); 
    return $str; 

    } 
    function compare($db_str, $app_str) 
    { 
     // $salt = "ol2ujh354238095uwef"; 
     $app_str = $app_str/*.$salt*/; 
     //echo "<br>Database password:- ".$db_str; 
     // echo "<br>Database password:- ".md5($app_str);    
     if(md5($app_str)==$db_str) 
     { 
      $response['success'] = '1'; 
     } 
     else 
     { 
      $response['success'] = '0'; 
     } 
     //echo json_encode($response); 
     //$response = json_encode($response); 
     die(json_encode($response)); 
     //mysqli_close($con); 
    }    
} 
    $func = new functions(); 
    $func->getUser(); 

?> 

Answer 1

username = userName.getText().toString(); 
pass = password.getText().toString(); 

你应该在onClick方法上做，而不是onCreate。

Answer 2

那么你收到此字符串：

[ 08-08  23:31:46.812 17362:17472 W/System.err ] 

从该服务器不是有效的JSON。 此行：

JSONObject responseObject = new JSONObject(str); 

试图将该字符串解析为json并失败。

我认为你总是期望从服务器的JSON响应，所以我怀疑你的这句话：

我的PHP脚本运行完美。

是正确的。你应该检查你的PHP服务器为什么会响应这个非JSON响应。