Can't catch variable passed

nin

This login page is supposed to bring you back to login page & display a error message if userid entered is invalid.
What i do is to pass a error message in hidden field regardless of success login or not. But only catch the error message in the login page. But don;t know y can;t catch the message.
Can someone help?? I greatly appreciate any help given. Thx.

<?php echo $_POST['error'];?>

<form id="form1" name="form1" method="POST" action="<?php echo $loginFormAction; ?>">
<table width="470" border="1" align="center">
<tr>
<td width="192"><strong>User ID</strong></td>
<td width="262"><label>
<input type="text" name="uid" id="uid" />
</label></td>
</tr>
<tr>
<td><strong>Password</strong></td>
<td><label>
<input type="password" name="pass" id="pass" />
</label></td>
</tr>
<tr>
<td> </td>
<td><label>
<input type="submit" name="register" id="register" value="Submit" />
<input name="error" type="hidden" id="error" value="Unknown User ID or password. Please try again." /> </label></td>
</tr>
</table>
</form>

YAOMK

You can start by including your hidden field inside the form, right now it is placed after the submit field.

I'm not sure what $loginFormAction is supposed to contain, but when you submit a form, those values are only available to the script you submitted them to. If you redirect the user to some other page, the data in $_POST gets flushed. If you are posting to the same script, there might be an error in the rest of your code.

nin

Actually, hidden field is inside the form. $loginFormAction is actually a dreamweaver generated server behaviour(login user).

chasiv

If you want $_POST['error'] to display at the top of the form (as you have it coded above), you should test to be sure that $loginFormAction is referrering to the same page you are on (so the form should be posting to itself). Try displaying the $loginFormAction variable to screen:

<?php echo "form action = " . $loginFormAction . "<br />";

Then I would change your
$loginFormAction
and instead return the entire $_POST variable array

<?php
ehco "<PRE>";
print_r($_POST);
echo "</PRE>";

to make sure you are getting everything you anticipate.

That should get you going on troubleshooting your form. If not, post the results of the page with the above results please

nin

Thx for everyone's help. I just realised tat the codeis illogical.

This is the correct code.

This is part of the code from the dreamweaver LogInUser server behaviour. Added that if login failed error variable will be set.
if (isset($_POST['uid'])) {
$MM_redirectLoginFailed = "login.php?error=1";

That's to be written @ the body part. Where ever you want the message to be displayed. Also, I have removed the hidden field as it's not needed at all.
<?php if(isset($_GET['error'])){ ?>
<p align="center" class="warning" >User name or password is invalid. Please login again. </p>
<?php } ?>

This is all corrected by my lecturer. She did mention y cannot use POST to catch the error. Because dreamweaver redirect login failed using
header("Location: ". $MM_redirectLoginFailed );
Said something about header only passed the URL. Not the values to the other page. That's what i faintly rmb. Sorry if explanation is wrong.

Hope this will help other newbies who's stuck @ dd login page with dreamweaver's login user server behaviour. with a little php coding.