[RESOLVED] How to extend this for multiple includes?

conjurer

Thanks to the help here I got this code working:

<?php 
require_once('includes/test.inc');
//set page specific variables
$pgtitle = "OMGMA Test Page";
$css = "../main.css";
do_html_header($pgtitle,$css);

?>
</head>
<body>

The do_html_header() function from the test.inc files is as follows:

<?php
function do_html_header($title, $style)
{
  // print an HTML header
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">
<html lang="en">
<head>
<title><?php echo $title; ?></title>
<link rel="stylesheet" type="text/css" href="<?php echo $style; ?>">

<?php
}		
?>

So what I am wanting to do now is to be able to pass more than one CSS file link to the function.

What would be the preferred method to do this? Do I need an array?

To be a little more specific I want to be able to put in my page, variables for more than one css file. I could see using up to 4 links, although sometimes I might only have one or two files. So assuming four it would look like this:

<?php 
require_once('includes/test.inc');
//set page specific variables
$pgtitle = "OMGMA Test Page";
$css1 = "../main.css";
$css2 =  "../other.css";
$css3 =  "../yetanother.css";
$css4 =  "../onemore.css";

do_html_header($pgtitle,$css1, $css2, $css3, $css4);

?>
</head>
<body>

So how could I then modify the do_html_header() so that it would print out a link to each of the variables but not print a link for variables that didn't exist? So if I specified two files I get two links, if I specify four I get four.

Would I do it like this? Or is there a better way?

<?php
function do_html_header($title, $style1, $style2, $style3, $style4)
{
  // print an HTML header
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">
<html lang="en">
<head>
<title><?php echo $title; ?></title>
<?php
   if $style1 {
        echo "<link rel='stylesheet' type='text/css' href='".$style1 ."'>";
  }
   if $style2 {
        echo "<link rel='stylesheet' type='text/css' href='".$style2 ."'>";
  }
  if $style3 {
        echo "<link rel='stylesheet' type='text/css' href='".$style3 ."'>";
  }
  if $style4 {
        echo "<link rel='stylesheet' type='text/css' href='".$style4 ."'>";
  }

//close of function:
}		
?>

matt72

You can do it this way, but it limits the amount of css files you can have to the number of arguments you put in the function declaration.

function do_html_header($title, $style1 = '', $style2 = '', $style3 = '', $style4 = '') 
{ 
  // print an HTML header 
?> 
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" 
"http://www.w3.org/TR/html4/loose.dtd"> 
<html lang="en"> 
<head> 
<title><?php echo $title; ?></title> 
<?php 
   if($style1 != '') { 
        echo "<link rel='stylesheet' type='text/css' href='".$style1 ."'>"; 
  } 
   if ($style2 != '' { 
        echo "<link rel='stylesheet' type='text/css' href='".$style2 ."'>"; 
  } 
  if($style3 != '') { 
        echo "<link rel='stylesheet' type='text/css' href='".$style3 ."'>"; 
  } 
  if($style4 != '') { 
        echo "<link rel='stylesheet' type='text/css' href='".$style4 ."'>"; 
  } 
//close of function: 
}

It's be much more succinct using an array and a loop

// in original file
$css = array();
$css[] = 'css1.css';
$css[] = 'css2.css'; // etc.

function do_html_header($title, $css) 
{ 
  // print an HTML header 
?> 
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" 
"http://www.w3.org/TR/html4/loose.dtd"> 
<html lang="en"> 
<head> 
<title><?php echo $title; ?></title> 
<?php 
foreach($css as $css_file)
{
        echo "<link rel='stylesheet' type='text/css' href='".$css_file."'>"; 
} 
//close of function: 
}

conjurer

Very Cool!

Very Very Cool!

Thanks!