dousui7410 2019-06-14 12:38
浏览 60

Ajax不断返回PHP的错误输出

this codes keeps returning 0 even if the username and password I entered is correct. I also did try the php codes without ajax but it works perfectly.

PHP

include('config/connection.php');

if(isset($_POST['login'])){
    $username = mysqli_real_escape_string($conn,$_POST['username']);
    $password = mysqli_real_escape_string($conn,$_POST['password']);

    $select_query = mysqli_query($conn,"SELECT * FROM tbl_users WHERE username='$username' AND password ='$password'");
    $count_rows = mysqli_num_rows($select_query);

    if($count_rows > 0){
        echo "1";
    } else {
        echo "0";
    }
}

AJAX

    $('form').on('submit', function (e) {

      e.preventDefault();

      $.ajax({
        type: 'post',
        url: 'phpscripts/function/fnc_login.php',
        data: $('form').serialize(),
        success: function (result) {
            //if fnc_login.php returned 1/true
            //redirect to index page
            if (result == 1) {                 
                window.location = "index.php";

            //if fnc_login.php returned 0/false
            } else {
                $('#message').html('User not found!').fadeIn('slow');
            }
        }
      });
    });
    return false;
  • 写回答

2条回答 默认 最新

  • dongwenhui8900 2019-06-14 12:44
    关注

    you check integer value in response not string

    if (result == "1") { }
    

    Hope it solve your problem.

    评论

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