Simple login form with php/javascript

Question

I'm working on a simple login in form that verfies one username and one password using php and javascript. I have a index.php a javascript file to check for errors and a results.php. I can't see anything that I'm doing wrong, but the page continues to submit Invalid login in no matter what is input. Looking for some hints if you guys can give me some. Thanks in advance. I have also looked at similar questions on here and my code pretty much matches the others.

results.php

<?php

$user="joesmith";
$password="abc123";
if(isset($_POST['submit'])){

if(($_POST['user'] == $user) && ($_POST['password'] == $password))
echo "You have successfully logged in";
}
else
{
echo "Invalid login! Try again.";
}
?>

index.php

<body>

<form class="basic-grey" action="results.php" method="post">

<div>
<label for="username" id="lUsernameLabel">* Username:</label>
<input type="text" name="username" id="iUsername"      onblur="validateUserName()" onfocus="resetUserName()" />
<div id="dUsernameError" class="errorMessage"></div>
</div>

<div>
<label for="password" id="lPasswordLabel">* Password:</label>
<input type="password" name="password" id="iPassword"  onblur="validatePassword()" onfocus="resetPassword()" />
<div id="dPasswordError" class="errorMessage"></div>
</div>

<div>
<input type="submit" class="button" value="Submit" />
</div>

Don't rely on JavaScript for input validation, checks must be done server-side, e.g. in PHP. — Mex
– Mex, Commented Feb 13, 2015 at 22:02
You can use your browser's developer tools to check what the actual outgoing HTTP request looks like. That would have saved you some time here. — Pointy
– Pointy, Commented Feb 13, 2015 at 22:02

Endacy · Accepted Answer · 2015-02-13 22:01:33Z

6

your input has name="username" but you are looking for $_POST['user']

answered Feb 13, 2015 at 22:01

Endacy

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7 Comments

lv0gun9 Over a year ago

Bull's eye! He found it!

Eric Smith Over a year ago

I'm confused, when the users puts a username in and hits submit is it not Posting that information? Or are you saying that the input needs to reflect $_POST as well?

Endacy Over a year ago

What I mean is your <input> element that is taking the username entered, is named 'username' in the html file, so it will be posted at $_POST['username'] to the php file, whereas you are looking for $_POST['user'] which doesn't exist.

Eric Smith Over a year ago

Oh okay I got you, though even changing it didn't make it succeed. On the results.php I changed if(($_POST['user'] ==$user) to if(($_POST['username'] == $user).

Endacy Over a year ago

Then I would suggest echo'ing out the two posted values just before they are compared.

|

Bilal Çınarlı · Accepted Answer · 2015-02-13 22:02:01Z

0

You are submitting username input with "username" but trying to check its validity with "user".

Change the part

($_POST['user'] == $user)

to

($_POST['username'] == $user)

it will start to work

answered Feb 13, 2015 at 22:02

Bilal Çınarlı

11 silver badge2 bronze badges

Comments

malcanso · Accepted Answer · 2015-02-13 22:02:23Z

0

Try $_POST['username'] instead of $_POST['user'].

(This would be a comment if I had the rep)

answered Feb 13, 2015 at 22:02

malcanso

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