Website Login mit Ajax

Question

Here is a screenshot of the console. Ich bin ein Anfänger in der Arbeit mit Json und Ajax und ich versuche, ein Login für meine Website zu schreiben, aber ich weiß nicht, warum es nicht funktioniert.

Login-Register.js

function loginAjax(){ 
$.post("login.js", function(data) { 
     if(data == 1){ 
      window.location.replace("/sportime.html");    
     } else { 
      shakeModal(); 
     } 
    }); 

} 

login.js

$(document).ready(function() 
{ 

    $('#login').click(function() 
    { 
    var username=$("#username").val(); 
    var password=$("#password").val(); 
    var dataString = 'username='+username+'&password='+password; 
    if($.trim(username).length>0 && $.trim(password).length>0) 
    { 
    $.ajax({ 
    type: "POST", 
    url: "ajaxLogin.php", 
    data: dataString, 
    cache: false, 
    beforeSend: function(){ $("#login").val('Connecting...');} 

    }); 

    } 
    return false; 
    }); 

    }); 

ajaxLogin.php

<?php 
include("conectare.php"); 
session_start(); 
if(isset($_POST['username']) && isset($_POST['password'])) 
{ 
// username and password sent from Form 
$username=mysqli_real_escape_string($mysqli,$_POST['username']); 
//Here converting passsword into MD5 encryption. 
$password=md5(mysqli_real_escape_string($mysqli,$_POST['password'])); 

$result=mysqli_query($mysqli,"SELECT uid FROM users WHERE   username='$username' and password='$password'"); 
$count=mysqli_num_rows($result); 
$row=mysqli_fetch_array($result,MYSQLI_ASSOC); 
// If result matched $username and $password, table row must be 1 row 
if($count==1) 
{ 
$_SESSION['login_user']=$row['uid']; //Storing user session value. 
echo $row['uid']; 
} 

} 
?> 

sportime.php

<?php 
    session_start(); 
    if(!empty($_SESSION['login_user'])) 
    { 
    header('Location: sportime-loggedin.php'); 
    } 
    ?> 

und dies ist der modale für die Anmeldung von sportime.php

<div class="modal fade login" id="loginModal"> 
       <div class="modal-dialog login animated"> 
        <div class="modal-content"> 
         <div class="modal-header"> 
          <button type="button" class="close" data-dismiss="modal" aria-hidden="true">&times;</button> 
          <h4 class="modal-title">Login</h4> 
          <div class="content text-center"><h5>Use your University credentials</h5></div> 
         </div> 
         <div class="modal-body"> 
          <div class="box"> 
           <div class="content"> 
            <div class="error"></div> 
            <div class="form loginBox"> 
             <form method="post" action="" onsubmit="loginAjax(); return false;" accept-charset="UTF-8"> 
              <input id="email" class="form-control" type="text" placeholder="University ID" name="id"> 
              <input id="password" class="form-control" type="password" placeholder="Password" name="password"> 
              <input class="btn btn-default btn-login" type="submit" value="Login"> 
             </form> 
            </div> 
           </div> 
          </div> 
         </div> 
        </div> 
       </div> 
      </div> 

Answer 1

Warum sind Sie post zu einer js-Datei versucht?

function loginAjax(){ 
$.post("login.js", function(data) { 
     if(data == 1){ 
      window.location.replace("/sportime.php");    
     } else { 
      shakeModal(); 
     } 
    }); 

} 

Ihre loginAjax Funktion wie dieses EDIT

function loginAjax() { 
    var username = $("#email").val(); 
    var password = $("#password").val(); 
    var dataString = 'username=' + username + '&password=' + password; 
    if ($.trim(username).length > 0 && $.trim(password).length > 0) { 
     $.ajax({ 
     type: "POST", 
     url: "ajaxLogin.php", 
     data: dataString, 
     cache: false, 
     beforeSend: function() { 
      $("#login").val('Connecting...'); 
     } 

     }).success(function(data) { 
     if (parseInt(data) > 1) { 
      window.location.replace("/sportime.html"); 
     } else { 
      shakeModal(); 
     } 
     }); 
    } 
    return false; 
    } 

aussehen sollte

Sie die IDs verlegt haben. Ersetzen

var username = $("#username").val();

mit var username = $("#email").val();