Concerning Include() and require(), the current php manual states, "The two constructs are identical in every way except how they handle failure."

However, I have read, and found online, "Unlike include(), require() will always read in the target file, even if the line it's on never executes. If you want to conditionally include a file, use include(). The conditional statement won't affect the require(). However, if the line on which the require() occurs is not executed, neither will any of the code in the target file be executed."

Does this second statement only concern earlier versions of PHP, or did I just not find it in the current manual?

My copy of the manual prepends those statements with:

"Prior to PHP 4.0.2, the following applies:"

According to Zeev Suraski (who probably should know these things, if anyone does), "as of PHP 4.0.3, the implementation of require() processes the file 'just in time'"
(http://marc.theaimsgroup.com/?l=php-...97419027907315)
i.e. only if needed. So, nowadays, the lines:

if(false){
    require("page.php");
}

Will neither read nor execute page.php. Ever. This is actually how include always behaved.

HTH
Dan

Dan - Thanks for clearing it up! I now see the "Prior to PHP 4.0.2" note in the require() documentation, but it wasn't quite clear to me the first time I read the note.

I did understand that include() always worked this way, but was reading a book which made me wonder whether I should go back and change some of my require() coding. I know the 2002 publication date of the book should have given me a clue, but sometimes out-of-date web "help" is harder to catch.

Thanks again!

- Steve