result.success在Ajax调用中失败

Question

我的代码运行良好，但在成功调用时失败。在ajax节“if（result.success）{”它显示浏览器源选项卡中的错误，结果为空。

登录页面：

<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> 

连接页面：

<?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点儿奇怪。不知道我去哪里错了任何帮助？

编辑：

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; 
      }; 

变化

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

result这里是服务器，而不是一个对象发送TE数据。

---

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

这是因为在你的UPDATE查询没有WHERE条款。

从文档：

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

编辑更新

后，你有一个额外};。

变化

success: function(result){ 

       window.location = "my-dashboard/"; 

      }; 
     }, 

到

success: function(result){ 

       window.location = "my-dashboard/"; 


     }, 

几个点值得一提：

• 弃用通知：jqXHR.success（），jqXHR.error（）和jqXHR.complete（）回调将在jQuery 1.8中被弃用。要准备代码以便最终删除它们，请改用jqXHR.done（），jqXHR.fail（）和jqXHR.always（）。

• mysql_ *函数被弃用。改用PDO。

Answer 2

就试试这个：

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

因为这个函数调用成功的结果。 第一个参数是来自服务器答案的数据。这不是对象。 你能在那里找到细节：http://api.jquery.com/jQuery.ajax/