Question

我有这个ajax代码。数据包含登录凭据。

$.ajax({
        url: "app/php/login.php",
        type: "GET",
        data: data,
        dataType: 'json',
        async: true,
        success: function(response){
            blah blah
        }
})

这是我的login.php。每当我发送请求时，都会弹出一条警告，其中显示消息 Something Went Wrong 。我这样做对吗？请原谅我在登录中使用GET方法

header('Content-Type: application/json');

$dbconn = mysqli_connect("localhost","root","","alumni_tracker") or die("Could not connect to database!"); //host, username, password, db
mysqli_select_db($dbconn,"alumni_tracker");
$student_no = $_GET["student_no"];
$password = $_GET["password"];
$query = "SELECT * FROM user WHERE student_no= '$student_no' AND password=MD5('$password')";
$res = mysqli_query($dbconn, $query);

if(empty($res)){
    $data = "1";
}
else if(!empty($res) && $student_no == "111111111"){
    $data = "2";
}
else{
    while($row = mysqli_fetch_array($res)){
        $data = array('student_no'=>$row['student_no'],'password'=>$row['password']);
    }
}
return json_encode($data); 
exit();

Answer 1

无视安全问题，修复方法是在 json_encode（$ data）

中使用 echo 而不是 return

PHP和Ajax出了点问题

1 个答案: