無法從JSON

Question

的價值，這是在PHP代碼

<?php session_start(); 

//Connect to database from here 
    $link = mysql_connect('****', '****', '****'); 
    if (!$link) { 
     die('Could not connect: ' . mysql_error()); 
    } 
    //select the database | Change the name of database from here 
    mysql_select_db('****'); 

//get the posted values 
$user_name=htmlspecialchars($_GET['user_name'],ENT_QUOTES); 
$pass=$_GET['password']; 

//now validating the username and password 
$sql="SELECT user_name, password FROM tbl_user WHERE user_name='".$user_name."'"; 
$result=mysql_query($sql); 
$row=mysql_fetch_array($result); 

//if username exists 
if(mysql_num_rows($result)>0) 
{ 
    //compare the password 
    if(strcmp($row['password'],$pass)==0) 
    { 
     // Return success message 
     $message = array("flag" => "1", "msg" => "Login successfully."); 
     echo json_encode($message); 
     //Regenerate session ID to prevent session fixation attacks 
     //session_regenerate_id(); 


     //now set the session from here if needed 
     //$_SESSION['u_name']=$user_name; 
     //$member=mysql_fetch_assoc($result); 
     //$_SESSION['u_id']=$member['id']; 
     //$name_show=$member['first_name'].' '.$member['last_name']; 
     //$_SESSION['name']=$name_show; 
      //Write session to disc 
      //session_write_close(); 

     } 
    else 
     // Return error message 
     $message = array("flag" => "0", "msg" => "Incorrect password."); 
     echo json_encode($message); 
} 
else //if username not exists 
{  // Return error message 
     $message = array("flag" => "0", "msg" => "Incorrect id."); 
     echo json_encode($message); 
} 
mysql_close($link);  
?> 

這是HTML代碼

    $.ajax({ 
       type: "get", 
       url: "http://mydomain.com/ajax_login.php", 
       data: { 
        user_name: $("#username").val(), 
        password: $("#password").val() 
       }, 
       success: function(jsondata){ 

        if (jsondata.flag == "0"){ 
         //if flag is 0, indicate login fail 
        } 
        else { 
         //if flag is 1, indicate login success and redirect to another page 
         window.location.href = 'ABC.html'; 
        }      
       }, 
       datatype: "json" 
      }); 

顯示屏顯示"{\"flag\":\"0\",\"msg\":\"Incorrect id.\"}"

我的問題是，現在連標誌是0，它仍然會轉到ABC.html 如果要修改if子句，那麼如果標誌爲0，它仍然會在子句的真實部分？

EDITED這是編碼

Answer 1

務必在您dataType: "json"配置您的Ajax調用jQuery的AJAX方法，並且一定要發送JSON頭在喜歡你的PHP迴應。

header("Cache-Control: no-cache, must-revalidate"); 
header("Expires: Sat, 26 Jul 1997 05:00:00 GMT"); 
header("Content-type: application/json"); 
echo json_encode($yourArray); 

如果你沒有正確配置你的ajax調用，結果可能被解釋爲一個簡單的字符串。

Answer 2

也許jsondata.flag是未定義的更多細節。
您需要解碼響應字符串以將其用作JS對象。
或者設置dataType : 'json' ...