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For 2023 - we are now fully retired from IT training.
We have made many, many friends over 25 years of teaching about Python, Tcl, Perl, PHP, Lua, Java, C and C++ - and MySQL, Linux and Solaris/SunOS too. Our training notes are now very much out of date, but due to upward compatability most of our examples remain operational and even relevant ad you are welcome to make us if them "as seen" and at your own risk.

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Posted by dabbler (leah), 18 February 2005
I'm modifying an existing page which was simply including the output from a perl script.

include ("");

Now I need to pass a HTTP_REFERER value through to it to find where the user first clicked the contact button, so I am doing so in the url of contactus.php?refer=, reading the file and trying to use preg_replace to put the value into the form in a hidden field using this in contactus.php:

$refer = $_GET['refer'];

$lines = file("");
foreach ($lines as $line)
     { $find="/xRef/";
           $replace = $refer;
           echo $line;

I don't get any errors, and it displays properly, but looking at the page source, it's not replacing the xRef value and I can't figure out why not, I can echo $refer at the top and can see the value is there. I expect it's just my first experience with preg_replace and I'm doing something silly.

Posted by admin (Graham Ellis), 18 February 2005
preg_replace RETURNS the modified string rather that changing it "in situ".  

Try replacing
   $line = preg_replace($find,$replace,$line);  

Posted by dabbler (leah), 18 February 2005
Ah-hah! That did the trick, thank you Graham!

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