An exceptionally easy way to do this would be to pass both $PHP_SELF and
$REQUEST_URI.
Alternatively, you can use explode():
<?php
$array = explode('?',$REQUEST_URI);
?>
<form method="post" action=<?php echo $array[0] ?>">
Username: <input type="text" name="username"><r>
etc</form>
-----Original Message-----
From: James, Yz [mailto:[EMAIL PROTECTED]]
Sent: Friday, February 16, 2001 2:57 PM
To: [EMAIL PROTECTED]
Subject: [PHP] Finding the? in $REQUEST_URI?page=blah
I really wouldn't ask this if I wasn't completely exausted, but the last
thing I feel like doing just now is reading up on this when someone can
probably answer it in a matter of seconds.....
OK, I've an error handling page. Basically, I want to split (that might
even be one of the functions I'll have to use) this sort of url:
error.php?/pages/login.php?username=name
into:
/pages/login.php
I'm using :
header("location: error.php?page=$REQUEST_URI");
When echoing the request_uri back on the error page, it omits the error.php?
bit, so I get:
login.php?username=name
The reason I want to do this is so that I can use a re-login form that will
submit to the page the user has badly logged into, so if they logged in to
main.php, the request_uri would be something like:
/pages/main.php?username=name
Which I'd want to rebuild on the error page, into a form, like:
<form method="post" action=<? echo "$REQUEST_URI"; /* Minus the suffix the
page may include */ ?>">
Username: <input type="text" name="username"><r>
etc</form>
Can anyone help?
Thankees :)
James.
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