Check if include exsists?

Want to supply error page if it doesn't.

[Sypher_5]

I use this working code to switch the file I include.

$foo = "$d";// foo is equal to the d varible 
if($foo == "") {// if foo is equal to nothing 
 $dir ="_pages";// then $dir is _pages 
} 
else {// else 
$dir = "$d";// $dir is equal to $d varible 
} 
$page = "$p.inc";// page is equal to the p varible 
if($page == ".inc") {// if page is equal to nothing 
 $content = "home.inc";// then $content is $dir/homes.inc 
$counter = "1";// & visitor counter is called 
} 
else {// else 
 $content = "$page";// $content is equal to $dir/$page varibles 
} 
// which means is the string is?d=_incl&p=page1 
// then $content should be page1.inc in the _incl 
// folder ( _incl/page1.inc )

Now I want to test that the include specified in the string exists, and if it doesn't that it includes an error page.

I've tried having a go with my novice skills but I'm getting parse errors.

if (is_file("$incl") {  
include($incl);  
}  
else { 
include(_incl/error.inc);  
}

Also tried using "file_exists" instead of "is_file".

On an added note is this going to cause much server slow down checking for the file?

I note that the pre and code tags here dont seem to work either..

[edited by: jatar_k at 4:19 pm (utc) on Jan. 14, 2004]
[edit reason] fixed sidescroll [/edit]

[coopster]

You are getting parse errors because you are missing a closing parenthesis. Try this:

if (is_file("$incl")) { // <-- note closing parenthesis! 
 include($incl);  
} else {  
 include('_incl/error.inc');  
} 

[Sypher_5]

Thanks, should've know it was a syntax error.