Question

这是php中的代码

    <?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);     
?>

这是html中的代码

                  $.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"
            });

显示屏显示"{\"flag\":\"0\",\"msg\":\"Incorrect id.\"}"

我的问题是，现在即使旗帜为0，它仍然会转到ABC.html if子句应该被修改为什么，如果该标志为0，它仍将在该子句的真实部分？

编辑这是编码的更多细节

Answer 1

或许jsondata.flag未定义您需要解码响应字符串以将其用作JS对象或者设置dataType : 'json' ...

Answer 2

请务必在jQuery ajax方法中使用dataType: "json"配置ajax调用，并确保在PHP响应中发送JSON标头，如。

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);

如果你没有正确配置你的ajax调用，结果可能被解释为一个简单的字符串。

无法从JSON获取值

2 个答案: