Include Filesnames with variables

nainil

Hi,

I am trying to make hyperlinks with the help of variable names.

I want something like ..
If $domainname= mydomainname.com
My code is something like: http://mydomainname.com/variables.php

// index.php

<?php

// directly finds current URL
$domainname = "http://$HTTP_HOST/";
echo $domainname;
?>

<?php include("$domainname.variables.php"); ?>

<?php include("$domainname.header-meta-tags.php"); ?>

<body link="#4D5AAC" vlink="#4D5AAC" alink="#4D5AAC" leftmargin="0" topmargin="0" marginwidth="0" marginheight="0">
<?php include("$domainname.top.php"); ?>
<?php include("$domainname.left.php"); ?>
</td>
<td width="2" bgcolor="#4D5AAC">
<img src="<?php echo $domainname."images/transparent.gif"; ?> width="2" height="1">
</td>

But this script doesnot work.. It doesnot give the URL in the INCLUDE AHREF LINK. How do i do it?

Thank you.
Nainil Chheda.

devinemke

it looks like you are trying to concatenate variables in your includes but your syntax is slightly wrong.

$domainname = 'http://' . $_SERVER['HTTP_HOST'] . '/';

include($domainname . 'variables.php');
include($domainname . 'top.php');
include($domainname . 'left.php');

nainil

I am running the script in a directory.

It doesnot work? I have links on the other pages. They show something like mydomainname.com/directoryname/page1.php

But i want it to be just mydomainname.com/page1.php

Thanks.

Nainil.

nainil

Hi,

I am trying to make hyperlinks with the help of variable names.

I am running the script in http://mydomainname.com/directory1/index.php

At present when I use the included filenames it gives me problems like the links which r there in the file, they show the current links. like..:
The included file : Left.php it is located in the main web folder.
It has links.

These links open like http://mydomainname.com/directory1/linkname.php
But they should look like:

I want something like ..
If $domainname= mydomainname.com
My code is something like: http://mydomainname.com/linkname.php

// index.php

<?php

// directly finds current URL
$domainname = "http://$HTTP_HOST/";
echo $domainname;
?>

<?php include("$domainname.linkname.php"); ?>

<?php include("$domainname.header-meta-tags.php"); ?>

But this script doesnot work.. It doesnot give the URL in the INCLUDE AHREF LINK. How do i do it?

Thank you.
Nainil Chheda.