preg_replace() not working.....parameters are wrong HELP!

[benjy]

ok i have this in one of my php scripts:

$template_1 = include("template_1.txt");
$keyword_url = "%%url%%" ;
$replacement_url = "$url";
$template_1 = preg_replace("$keyword_url", "$replacement_url");

in the file "template_1.txt" there is a short html code, in that code somewhere there will be "%%url%%" what i want to do is get preg_replace to get the variable $url which contains html to replace %%url%% with $url which would have been submitted from a external form. but i keep geting this error:

Warning: Wrong parameter count for preg_replace() in /home/comesept/public_html/introverted/ll-url.php on line 41

which parameters are incorect? how would i reolve this issue?

[xblue]

why not give the manual a quick look?

preg_replace takes 3 argument, pattern, replacement, and the string to work on, which should probably be the content of template_1.txt.

however, include() doesn't return the content of the file, it simply includes it in your code as if you had pasted it at that position. check file() or fread() if you want to get the content into a variable.

you also seem to have some trouble using the pattern delimiters correctly, I'd recommend using str_replace() for simple replacements like this one.