PHP腳本上的jQuery.ajax POST 500內部服務器錯誤

Question

我在將ajax請求發送到登錄腳本的js函數上遇到此錯誤。 這是代碼：

function validateLogin(){
  $(document).ready(function(){
    $("#loginform").submit(function(){
      $.ajax({
        type: "POST",
        url: "./php/autologin.php",
        data: {
          'login': $("#login").val(), 
          'password': $("#password").val()
        },
        dataType: "json",
        success:function(data){
          if(data.status === "success"){
            alert(data.message);
            window.location = "./userarea.html?user=".concat(data.user);
          } else if (data.status === "error"){
            alert(data.message);
          }
        },
        error:function(thrownError){
          console.log(thrownError);
        }
      }); 
      return false;
    });

  });
}

我從上述代碼的URL上收到POST 500 Internal Server Error，這是chrome控制台中正在打印的內容：

POST http://localhost/capstone/atomos-4.0/php/autologin.php 500 (Internal Server Error) jquery-1.7.2.min.js:4
f.support.ajax.f.ajaxTransport.send jquery-1.7.2.min.js:4
f.extend.ajax jquery-1.7.2.min.js:4
(anonymous function) validateLogin.js:4
f.event.dispatch jquery-1.7.2.min.js:3
f.event.add.h.handle.i jquery-1.7.2.min.js:3

我不太了解發帖錯誤要告訴我的內容，任何人都可以向我提出一些正確的方向來解決此問題。 讓我知道您是否需要更多信息。 謝謝

編輯：這是有問題的php函數： autologin.php

include './login.php';

$login = $_POST['login'];
$password = $_POST['password'];
$loginattempt = json_decode(login($login, $password));

if ($loginattempt->{'status'} === "success") {
  return json_encode(array('status' => "success", 'message' => "Login Successful!"));
  die();
} else { 
  return json_encode(array('status' => "error", 'error' => "loginfailure", 'message' => ($loginattempt -> {"message"})));
  die();
}
?>

和login.php ：

<?php

include './connnect_to_mysql.php';

function login($log, $pass) {
  $connection = json_decode(connect_to_mysql());
  if ($connection->{'status'} === "success") {
    $sqlquery = mysql_query("SELECT * FROM userdata WHERE login='$log' AND password='$pass'") or die(mysql_error());
    if (mysql_num_rows($sqlquery) == 1) {
      setcookie("user", $log, 60 * 60 * 24);
      return json_encode(array('status' => "success", 'message' => "Login Successful.", 'user' => $log));
      die();
    } else {
      return json_encode(array('status' => "error", 'error' => "loginfailure", 'message' => mysql_error()));
      die();
    }
  } else {
    return json_encode(array('status' => "error", 'error' => "connectionerror", 'message' => $connection -> {'message'}));
    die();
  }
}

?>

編輯2： connect_to_mysql.php ：

<?php

function connect_to_mysql() {
  $con = mysql_connect('localhost', 'user', 'pass') or die(mysql_error());
  if ($con) {
    $db = mysqli_select_db($con, 'db')or die(mysql_error());
    if ($db) {
      return json_encode(array('status' => "success", 'message' => "connected"));
      die();
    }
  } else {
    return json_encode(array('status' => "error", 'message' => mysql_error()));
    die();
  }
}

?>

Answer 1

修復include './connnect_to_mysql.php'; include './connect_to_mysql.php'; :)