How Do I Insert a Variable into a PHP 'Include'?

Question

Not much to explain really, I just cant figure out the correct syntax to use a variable in an include URL

heres what I thought it would be:

$round_num=1;
include '../../fixtures/round{$round_num}fix.php';

It returns the usual:

Warning: include(../../fixtures/round{$round_num}fix.php) [function.include]: failed to open stream: No such file or directory

This must be a very simple one for allot of you out there lol but i just cant find my answer ANYWHERE!

alex · Accepted Answer · 2012-11-25 06:50:33Z

3

Use double quotes (") to delimit your string. String interpolation doesn't occur in single quoted (') strings.

$round_num=1;
include "../../fixtures/round{$round_num}fix.php";

answered Nov 25, 2012 at 6:50

alex

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2 Comments

Thank you so much! so i have to use double quotes whenever i want to use a variable in something like this?

@ProdigyDoo Or concatenation.

JP_ · Accepted Answer · 2012-11-25 06:55:59Z

0

you can also use alternate syntax such as:

include "../../fixtures/round" . $round_num . "fix.php";

Sometimes people prefer to have the variable highlighted by their IDE for clarity.

answered Nov 25, 2012 at 6:55

JP_

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Personally I don't see the point of using double quotes and escaping the variable.

I usually don't do this;however, it's very common.