Parapluie Parapluie - 2 months ago 15
PHP Question

avoid duplication of php include/require

I have been wondering this for a while, and I have not yet seen the answer in SE.

When using includes in php, my working understanding is that any code that is included will be processed by the server.

This would mean that in the following conditional, both large blocks of code are PROCESSED on the server even though only one of these includes will be EMPLOYED.

<?php
if($conditional === TRUE){
include("MASSIVEAMOUNTOFCODE.php");
}
else {
include("GARGANTUANAMOUNTOFCODE.php");
}
?>


(Please pardon the grossly simplified code here.)

I have recently started using the following structure to perform the same task, with the aim of saving processing time on the server. Please tell me if this loads only a single include file, or if I am on a fool's errand:

<?php
if($conditional === TRUE){
$includeFile = "MASSIVEAMOUNTOFCODE.php";
}
else {
$includeFile = "GARGANTUANAMOUNTOFCODE.php";
}

include("$includeFile");
?>


Note the double usage of "include" in the first sample, and only a single use of "include" in the second sample.

Answer

Your original assumption that both files are processed is wrong. An include statement inside a code block that's never executed is ignored.

You can see that this is true by putting a syntax error in the include file in the non-executed branch. I did:

testinclude.php:

<?php
if (false) 
    include 'badinclude.php';

echo "done\n";

badlinclude.php:

<?php

$foo =;

When I executed testinclude.php there was no complaint about the syntax error in badlinclude.php. But if I change if (false) to if (true) I get an error.

Parse error: syntax error, unexpected ';' in /Users/barmar/badinclude.php on line 3

So the two versions of your code should be equivalent.