How to pass arguments to an included file

0 votes

I'm trying to make the whole <head> section its own include file. One drawback is the title and description and keyword will be the same; I can't figure out how to pass arguments to the include file.

So here is the code:

index.php

<?php include("header.php?header=aaaaaaaaaaaaaaaaaaaaa"); ?>

<body>
.....
..
.

header.php

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<link rel="shortcut icon" href="favicon.ico">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<meta name="Keywords" content=" <?php $_GET["header"]?> " >
<meta name="Description" content=" <?php $_GET["header"]?> " >
<title> <?php $_GET["header"]?> </title>
<link rel="stylesheet" type="text/css" href="reset.css" />
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.3/jquery.min.js"></script>
</head>

Obviously this doesn't work; how can I pass arguments to an included file?

Nov 20, 2020 in PHP by kartik
 • 37,510 points • 2,714 views

1 answer to this question.

0 votes

Hello @kartik,

Include has the scope of the line it's called from.

If you don't want to create new global variables, you can wrap include() with a function:

function includeHeader($title) {
    include("inc/header.php");
}

$title will be defined in the included code whenever you call includeHeader with a value, for example includeHeader('My Fancy Title').

If you want to pass more than one variable you can always pass an array instead of a string.

answered Nov 20, 2020 by Niroj
 • 82,880 points

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