[PHP] Re: PHP] can't pass complete URL (part of the query string)

[Nicole Lallande]

Chris -
Tried that -- I have urlencode in the script that sends the url and I have url encode right below the form action -- ref is getting cut off at the first ampersand - regardless:

where:
<form
action="http://embitec.com/fishcart/email.php?ref=http://embitec.com/fishcart/displayem.php3?cat=5&olimit=0&zid=1&lid=1";
method="post">
and:

<input type="hidden" name="ref" value"<? php echo rawurlencode($ref); ?>">
or even
<input type="hidden" name="ref" value"<? php echo rawurlencode($_GET['ref']); ?>">

yields:

<input type="hidden" name="ref" value="http%3A%2F%2Fembitec.com%2Ffishcart%2Fdisplayem.php3%3Fcat%3D4">


still cutting off everything after the ampersand..??

Best,

Nicole

<form
action="http://embitec.com/fishcart/email.php?ref=http://embitec.com/fishcart/displayem.php3?cat=5&olimit=0&zid=1&lid=1";
method="post">

There is your problem right there. Here are the variables you are
passing:

ref=http://embitec.com/fishcart/displayem.php3?cat=5
olimit=0
zid=1
lid=1

The URL you want to set ref to needs to be URL encoded. You can use
rawurlencode() to achieve this. You will know you have it right when
your HTML form tag looks like this:

<form
action="http://embitec.com/fishcart/email.php?ref=http%3A%2F%2Fembitec.com%2Ffishcart%2Fdisplayem.php3%3Fcat%3D5%26olimit%3D0%26zid%3D1%26lid%3D1";
method="post">

Hope that helps.

Chris

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Nicole Lallande
[EMAIL PROTECTED]
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