PHP include syntax

can I put a variable inside an include statement?

[foow]

Hi, please forgive this newbie question...

I'm using the following include

<?php include 'includes/I_searchbox.txt';?>

which is fine, but I want to include a variable ($ty_dir) in the include but I cant figure out the correct syntax to make it work

ie say $ty_dir = 'http://www.mysite.com/thisyearsdirectory/';
then I want PHP include that calls
[mysite.com...]
I've tried

<?php include $ty_dir'includes/I_searchbox.txt';?>

and others but can't get it to work. :(
Any help appreciated

[edited by: jatar_k at 5:51 pm (utc) on Jan. 4, 2004]
[edit reason] fixed the bbcodes [/edit]

[ikbenhet1]

<?php include ( 'includes/I_searchbox.txt?ty_dir='.$ty_dir )?>

This should work. I usually do like this:

$a='includes/I_searchbox.txt?ty_dir='.$ty_dir;
include($a);

[foow]

thanks kbenhet1, that example didn't seem to work for me but learning from it I tried this

<?php include ( $ty_dir.'includes/I_searchbox.txt' )?> 

and it works! - thanks a million!

[ergophobe]

foow,

In case you don't know why... Your last version works because you have put a concatenation operator (.) between your variable and your literal string. That is what makes it work.

One good thing to do if your include doesn't work is to echo it (replace 'include' with 'echo') and see if the string it prints is the name of the file or not.