result.success在Ajax調用中失敗

Question

我的代碼運行良好，但在成功調用時失敗。在ajax節「if（result.success）{」它顯示瀏覽器源選項卡中的錯誤，結果爲空。

登錄頁面：

<form id="login_form_header"> 
    <input type="hidden" name="action" value="sc_ajax_callback" /> 
    <input type="hidden" name="func" value="sc_login" /> 

    Email: 
    <input type="text" name="username" /> 

    Password: 
    <input type="password" name="password"/> 

    <input type="submit" value="Log in" /> 
</form> 
<script> 
$("#login_form_header").submit(function(event){ 

    event.preventDefault(); 

    $.ajax({ 
     type: 'post', 
     url: '<?php echo get_stylesheet_directory_uri(); ?>/connect.php', 
     data: $('#login_form_header').serialize(), 
     dataType: 'json', 
     success: function(result){ 
      if (result.success){ 
       window.location = "my-dashboard/"; 
       return false; 
      }; 
     }, 
     error: function(e){console.log("Could not retrieve login information")} 
    }); 

    return false; 
}); 
</script> 

連接頁面：

<?php 

    mysql_connect("localhost", "%user%", "%pass%") or die(mysql_error()); // Connect to database server(localhost) with username and password. 
    mysql_select_db("%db%") or die(mysql_error()); // Select registration database. 
    # Make sure form data was passed to the script 
    if(isset($_POST['username']) && !empty($_POST['username']) AND isset($_POST['password']) && !empty($_POST['password'])){ 

    # Define Variables 
    $active_var = "1"; 
    $given_username = mysql_escape_string($_POST['username']); 
    $given_password = dosomethingtopass; 
    $matched_username = ""; 
    $matched_password = ""; 

    # See if there is matching info in the database 
    $sql = 'SELECT username, password, active FROM %table% WHERE username="'.$given_username.'" AND password="'.$given_password.'" AND active="'.$active_var.'"'; 
    $result = mysql_query($sql); 
    while($row = mysql_fetch_assoc($result)){ 
    if($given_password == $row['password']){ 
     $matched_username = $row['username']; 
     $matched_password = $row['password']; 
     } 
    }; 

    # If there was a match 
    if($matched_username != "" && $matched_password != ""){ 
     #If there is only one result returned 
     echo json_encode(array("$matched_password")); 
     $session_sql = 'SELECT id, username, password, last_login FROM %table% WHERE username="'.$matched_username.'" AND password="'.$matched_password.'"'; 
     $session_result = mysql_query($session_sql); 
     while($returned_row = mysql_fetch_assoc($session_result)){ 
      if($matched_password == $returned_row['password']){ 


      $_SESSION['id'] = $returned_row['id']; 
      //$_SESSION['last_login'] = $returned_row['last_login']; 
      $_SESSION['username'] = $returned_row['username']; 
       } 
      } 

      $date = date('Y-m-d H:i:s'); 
      $update_sql = "UPDATE %table% SET last_login='".$date."'"; 
      mysql_query($update_sql); 

     echo json_encode(array("success"=>"user logged in", "session"=>$_SESSION)); 
     } 
    } 

?> 

另外，update_sql正在更新在我的表中的每一行有點兒奇怪。不知道我去哪裏錯了任何幫助？

編輯：

更新

$.ajax({ 
     type: 'post', 
     url: '<?php echo get_stylesheet_directory_uri(); ?>/connect.php', 
     data: $('#login_form_header').serialize(), 
     dataType: 'json', 
     success: function(result){ 

       window.location = "my-dashboard/"; 

      }; 
     }, 
     error: function(e){console.log("Could not retrieve login information")} 
    }); 

Answer 1

success: function(result){ 
      if (result.success){ 
       window.location = "my-dashboard/"; 
       return false; 
      }; 

變化

success: function(result){      
        window.location = "my-dashboard/";      
       } 

result這裏是服務器，而不是一個對象發送TE數據。

---

Also the update_sql is updating every row in my table kinda odd

這是因爲在你的UPDATE查詢沒有WHERE條款。

從文檔：

The WHERE clause, if given, specifies the conditions that identify which rows to update. With no WHERE clause, all rows are updated.

更多信息here

編輯更新

後，你有一個額外};。

變化

success: function(result){ 

       window.location = "my-dashboard/"; 

      }; 
     }, 

到

success: function(result){ 

       window.location = "my-dashboard/"; 


     }, 

幾個點值得一提：

• 棄用通知：jqXHR.success（），jqXHR.error（）和jqXHR.complete（）回調將在jQuery 1.8中被棄用。要準備代碼以便最終刪除它們，請改用jqXHR.done（），jqXHR.fail（）和jqXHR.always（）。

• mysql_ *函數被棄用。改用PDO。

Answer 2

就試試這個：

success: function(result){ window.location = "my-dashboard/"; }

因爲這個函數調用成功的結果。 第一個參數是來自服務器答案的數據。這不是對象。 你能在那裏找到細節：http://api.jquery.com/jQuery.ajax/