PHP Code Notice: Undefined variable: password in C:\xampp\htdocs\myfiles\login.php on line 51 you must enter your password?

Question

The question I am asking has been asked here but was asked pretty badly and resulted in the problem not being resolved. What the guy that asked that question is experiencing is exactly what I am experiencing, I assume he is following the same YouTube tutorial as me. Anyway here is all of the PHP / HTML code I have written to far:

<?php
session_start();
?>

<?php
$form = "<form action='./login.php' method='post'>
<table>
<tr>
<td>Username:</td>
<td><input type='text' name='user' /></td>
</tr>

<td>Password:</td>
<td><input type='password' name='password' /></td>
</tr>

<td></td>
<td><input type='submit' name='loginbtn'  value='Login'/></td>
</tr>
</table>

</form>";

if ($_POST ['loginbtn']){
$user = $_POST['user'];
$user = $_POST['password'];


if ($user){ 
if ($password){ 
echo "$user - $password <hr /> $form";
}
else
echo "you must enter your password $form";

}
else
echo "you must enter your username $form";

}
else
echo $form;

?>

I've tried adding a variable to if ($user){ and if ($password){ but that did nothing. Anyone know what the issue could be?

except for being a messy code, what error are you getting and what are you trying to achieve exactly? — dev7, Jan 27 '14 at 20:23
Sidenote: Don't store passwords in plain text. And I fear the rest of your code (SQL) is open to [SQL injection](http://stackoverflow.com/q/60174/) - You "are" using a DB right? Oh, still storing in `.txt`? ***Lovely*** — Funk Forty Niner, Jan 27 '14 at 20:32
And change `$user = $_POST['password'];` to `$password = $_POST['password'];` if you still want to make this work. — Funk Forty Niner, Jan 27 '14 at 20:33
Yea... I tried watching it and it made my brain _melt_ @PeeHaa — qwertynl, Jan 27 '14 at 20:33

score 1 · Accepted Answer · answered Jan 27 '14 at 20:23

1

$user = $_POST['user'];
$user = $_POST['password'];

$password is not properly defined, and you are using it in the if below

answered Jan 27 '14 at 20:23

cornelb

6,046
3
19
30

score 1 · Answer 2 · edited Jan 27 '14 at 22:21

1

$user = $_POST['user'];
$user = $_POST['password'];

This... you are assigning both user and password to the same variable ($user). That is why you have undefined variable password.

Also use htmlentities to prevent code injection:

$user = htmlentities($_POST['user']);
$password = htmlentities($_POST['password']);

edited Jan 27 '14 at 22:21

cssyphus

37,875
18
96
111

answered Jan 27 '14 at 20:30

score 0 · Answer 3 · answered Jan 27 '14 at 20:25

Sorry, that doesn't look like a very good tutorial. For one thing, I see that you create loginID and password fields on the page, but at what point is a user submit received? And where is the code for those values to be passed to PHP, evaluated, and returned?

I suggest these other tutorials instead:

phpAcademy - Register and Login - OOP version
phpAcademy - Register and Login - Procedural version

theNewBoston - Project Lisa (adv)

From your code, I think you will be most comfortable with the procedural tutorial from phpAcademy, but you should really learn the OOP one.

Other comments:

(removed b/c cornelb beat me to the punch -- see his answer.)

Thank you so much for the tutorial links. They look ten times better than the tutorial I'm following. — Roy Smith, Jan 27 '14 at 20:50

PHP Code Notice: Undefined variable: password in C:\xampp\htdocs\myfiles\login.php on line 51 you must enter your password?

3 Answers3