My First Login Form! ~ Still need help.

phpMyTony

Okay, so I have created this login form, and have learned how it works. The code I used is:

<html>
<head />
<body>

<form action="index.php" method=POST>
Username: <input type=text name=user><br />
Password: <input type=password name=pass><br />
<input type=submit value="Login!"><p>
</form>

<?php

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

if (($user=="Admin") && ($pass=="pass")) echo "Correct!";
else echo "Incorrect!";

?>

</body>
</html>

The demo is at: http://www.tonyperez.biz/login/index.php

Now, my question is, how can I make it so that it doesn't automatically show Incorrect!

Thanks 😉

mp1234

Try putting curly brackets for your control statement like so:

if (($user=="Admin") && ($pass=="pass")) {echo "Correct!";}
else {echo "Incorrect!";}

hope it helped...

phpMyTony

Hey MP, thanks for the reply.

It did not seem to make any difference :S

Krik

Try this

<?php
$status = '';
if(isset($_POST['user'])) {

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

if (($user=="Admin") && ($pass=="pass")) {
echo "Correct!";
}
else {
$status = "Incorrect!";
}
}
else {
?>
<html>
<body>
<?php echo $status; ?><br />
<form action="index.php" method=POST>
Username: <input type=text name=user><br />
Password: <input type=password name=pass><br />
<input type=submit value="Login!">
</form>
</body>
</html>
<?php
}
?>

phpMyTony

Ah, perfect thanks! Please can you explain how you done this, and what that code done?

Krik

Its real simple.

The $POST[] super global array is empty until your form posts data to it so what I did was check if the "user" key had a value.

if(isset($_POST['user']))

And if it was set it process the posted data if it was not set it displyed the form.

Now of course you needed it to display an error if the data submitted was wrong. And of course if it is wrong they will want to try and resubmit. So if the data was not correct it would set the "$status variable to "Incorrect" and display it with the form again.

Hmm, actually there is slight error in how I made that.

<?php
$status = '';
if(isset($_POST['user'])) {

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

if (($user=="Admin") && ($pass=="pass")) {
echo "Correct!";
}
else {
$status = "Incorrect!";
}
}
if((!isset($_POST['user'])) || ($status != '')) {
?>
<html>
<body>
<?php echo $status; ?><br />
<form action="index.php" method=POST>
Username: <input type=text name=user><br />
Password: <input type=password name=pass><br />
<input type=submit value="Login!">
</form>
</body>
</html>
<?php
}
?>

There that is better.

Notice I just changed the else to an if that checks whether if $_POST['user'] is not set or if $status is not ''. So if either is true it will show the form.

phpMyTony

Ah, I kind of understand a bit more. I must admit, some parts I don't understand, but thanks =]

VelvetElves

This is my first post, because I think I can help, in a way that is understandable.

<?php

$user=$POST['user'];
$pass=$POST['pass'];

if (($user=="Admin") && ($pass=="pass")) echo "Correct!";
if else (($user=="") && ($pass=="")) echo "";
else echo "Incorrect!";

I think this would work, because if nothing is there "" than display nothing ""

-Matt

PS: Sorry for my aweful grammer, hopefully I can pass 8th grade Engish lol.