AJAX > PHP Log in not working - php

Question

I'm trying to create a log in form using html > ajax > php, the problem is the php is not working, I don't know where is the problem, I think the ajax cannot execute my php file. I need help. Thanks in advance.
Here is my HTML code: my form and inputs are below
<form id="loginForm">
<input type="text" data-clear-btn="true" name="username" id="username" value="" placeholder="Username / ID No.">
<input type="password" data-clear-btn="true" name="password" id="password" value="" placeholder="Password">
<input type="checkbox" name="rem_user" id="rem_user" data-mini="true">
<label for="rem_user">Remember me</label>
<input type="submit" name="login" id="login" value="Log in" class="ui-btn" />
</form>
<div class="err" id="add_err"></div>
AJAX script that sends request on my php file
<script>
$(document).ready(function(){
$("#loginForm").submit(function(){
var username = $("#username").val();
var password = $("#password").val();
// Returns successful data submission message when the entered information is in database.
var dataString = 'username=' + username + '&password=' + password;
if (username == '' || password == ''){
alert("Please Fill All Fields");
}
else {
// AJAX Code To Submit Form.
$.ajax({
type: "POST",
url: "php/login-action.php",
data: dataString,
success: function(result){
window.location="#report_page";
}
});
}
return false;
});
});
</script>
PHP File
<?php
require "includes/connection.php";
include "includes/function.php";
if(isset($_POST['login'])){
$username = $_POST['username'];
$password = $_POST['password'];
$username = sanitize($username);
$password = sanitize($password);
$pass2 = md5($password);
$salt = "sometext";
$validateHash = $salt.$pass2;
$pass = hash("sha512", $validateHash);
$sql = "SELECT * FROM user_login WHERE username='".$username."' and password='".$password."'";
$result = mysqli_query($con,$sql) or die("Error: ". mysqli_error($con));
$count=mysqli_num_rows($result);
while($row=mysqli_fetch_array($result))
{
$id = $row['user_id'];
$username = $row['username'];
$name = "".$row['firstname']." ".$row['lastname']."";
$acc_type = $row['Acc_Type'];
}
if($count==1){
if($acc_type == 'user') {
$_SESSION["id"] = $id;
$_SESSION["username"] = $username;
$_SESSION["name"] = $name;
echo 'true';
}
else {
echo 'false';
}
}
}
?>

Answer 1

as Cattla mentioned in comments.
Your PHP is looking for $_POST['login'], and your $.ajax call didn't pass that in.
so here is the answer
var dataString = 'login=login&username=' + username + '&password=' + password;
Debug tips
Did ajax send all required inputs to PHP (you can inspect this from browser developer tool)
Did php receive all required inputs (you could var_dump($_POST)
Did php connect to mysql successfully
Did ajax receive data from PHP (use alert or console.log)

Answer 2

try this, and if you get error state what it is
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("#loginForm").submit(function(){
if (username == ' ' || password == ' '){
alert("Please Fill All Fields");
}
else {
// AJAX Code To Submit Form.
$.ajax({
type: "POST",
url: "php/login-action.php",
data: $(this).serialize(),
success: function(result){
alert('sucess'); //window.location="#report_page";
}
});
}
return false;
});
});
</script>