Question

我有一个使用ajax和php代码的登录表单。问题是始终返回错误而不是将我登录到系统中。我花了好几天试图找到错误，但我不能。

php：

<?php 
  include 'db.php';

  $email    = trim($_POST['email']);
  $password = trim($_POST['password']);

  $cek    = mysqli_query($conn, "SELECT * FROM user_csr WHERE email='$email' AND csr_pwd='$password'");    
  if(mysqli_num_rows($cek)>0)
  {
    echo 'true';
  }
  else
  {
    echo 'false';
  } 

?>

ajax：

function ceklogin(){
    var email    = document.getElementById('mail').value;
    var password = document.getElementById('pass').value;

        $.ajax({
            url: 'tes.php',
            method: 'POST',
            data: {email: email, password: password},
            success: function(html) {

                if(html == 'true')
                {
                     alert("login success"); 
                }
                else
                {
                     alert("login failed");
                }
            }
        });
}

<form>
    <input type="email" name="email" id="mail" required>
    <input type="password" name="password" id="pass" required>
    <button type="submit" class="w3ls-cart" onclick="ceklogin()">Sign In</button>
</form>

警报的结果是登录失败＆＃39;。但是电子邮件和密码都与数据库一致。希望有人能帮助我解决这个问题，提前谢谢。

Answer 1

这应该有效。只需确保您拥有下面标识的DIV即可显示结果。

function ceklogin() {
  var email    = document.getElementById('mail').value;
  var password = document.getElementById('pass').value;
  var xhttp = new XMLHttpRequest();
  xhttp.onreadystatechange = function() {
    if (this.readyState == 4 && this.status == 200) {
      document.getElementById("resultDiv").innerHTML = this.responseText;
    }
  };
  var sentInfo = "email=" + email + "&password=" + password;
  xhttp.open("POST", "YourPHPFileHERE.php", true);
  xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
  xhttp.send(sentInfo);
}

Answer 2

你应该尝试添加＆＃34; ===＆＃34;在你的if条件下。

          ` if(html === 'true')
            {
                 alert("login success"); 
            }
            else
            {
                 alert("login failed");
            }`

AJAX：PHP登录ajax确实有效

2 个答案: