simple variable php question

[zebono2]

im trying to call for a variable for part of a link

include ('$themedir; /index.php');

this is not working

the variable is $themedir = '/theme/default';

and is located in a diffent .php that had been included before

zebono2 Reviewed by zebono2 on 7th Jan 2012. simple variable php question im trying to call for a variable for part of a link include ('$themedir; /index.php'); this is not working the variable is $themedir = '/theme/default'; and is located in a diffent .php that had been included before Rating: 5

[hugo04]

1. can't call variable inside include
2. for your work, I suggest you do

if (index){
$themedir = '/theme/default';
}else{
$themedir = ' '/theme/default2';
}

load $themedir


hope this will help you guy

[soft2050]

1. can't call variable inside include

Wrong, you can use variables inside a include statement

----------------------------

include ('$themedir; /index.php'); 

Why is there a ; in the code and to use variables you need to use double quotes (")

include ("$themedir/index.php"); 

Also, the themedir variable:

$themedir = '/theme/default'; 

Whats the path from the script you are using?
use realpath() to check whether you are on the right directory or not

http://php.net/manual/en/function.realpath.php

[jokerhacker]

soft2050 is right. just another note, if you get problems about the path, use

include (getcwd() . "/$themedir/index.php"); 

[zebono2]

thanks guys i got it working by using

include ("$themedir/index.php");

[Robin H]

I'm actually not a big fan of double quotes.

Personally I tend to always use single quotes when possible.
Here you have both methods:

<?php
include($themedir . '/index.php'); // Notice the dot
?>

<?php
include("$themedir/index.php");
?>