How can I echo php with php?

Question

The website that I'm developing uses a "header" and "footer" template. These templates are included on each page of the site. The header template opens the HTML and contains the complete head tag, and then opens the body tag. The footer template closes the body tag and then closes the HTML. The basic structure is as follows:

<?php
//Set the current page name so it can be highlighted in the menu bar
$page="Home";
//For page-specific HEAD includes
$headincludes=array("<some type of code>","<some other type of code>");
include("/assets/template/header.php");
?>
<!--START PAGE CONTENT-->
<!--END PAGE CONTENT-->
<?php
include("/assets/template/footer.php");
?>

The header template dumps the "headincludes" array into the head section of the page as so:

foreach ($headincludes as $toecho)
echo $toecho;

This method works fine for things such as javascript, but when I try to add a PHP statement, it ends up as unparsed PHP in the final page served to the browser. Is there some special method to echoing PHP with PHP?

Could you show the example of how are you adding php statment in the $headincludes array? — Shakti Singh
– Shakti Singh, Commented Mar 7, 2011 at 6:34
Couldn't you just output php code without the <?php ?> tags? — sh03
– sh03, Commented Mar 7, 2011 at 11:36

Dan Lugg · Accepted Answer · 2011-03-07 06:49:55Z

2

Yes, eval() (read evil()) however you're best to reconsider your approach. Perhaps, if you're looking to include a considerable amount of PHP code, pass an array of files that could further be included:

// at the beginning of the file
$includes = array('db.php', 'config.php');


// in your header, or footer (whichever is applicable)
foreach($includes as $file){
    if(file_exists($file)){
        include $file; // or include_once
    }
}

Then, using the dynamic includes, you can manage and load additional code as necessary.

edited Mar 7, 2011 at 6:49

answered Mar 7, 2011 at 6:35

Dan Lugg

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1 Comment

sh03 Over a year ago

Mmm, no. evil() just execute php as string. It does not allow you to echo php.

alex · Accepted Answer · 2011-03-07 06:33:52Z

0

The way to execute PHP from a string is with eval(), however, it is rarely the right tool to use.

Can you place the PHP you need above $headincludes, and then add its output to $headincludes?

<?php
//Set the current page name so it can be highlighted in the menu bar
$page="Home";

include 'generators.php';

$links = generateLinks('some_arg');

//For page-specific HEAD includes
$headincludes=array("<some type of code>","<some other type of code>", $links);
include("/assets/template/header.php");
...

answered Mar 7, 2011 at 6:33

alex

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1 Comment

bnrup Over a year ago

I realized that it was far simpler to go with a different approach to structuring - by simply dividing things up into more modules. Now, I have a file for Global page code, a section for page-specific head code, a file for the page header and navigation, the body of the page, and a file for the footer. It's amazing how much a different perspective can change things.

j_freyre · Accepted Answer · 2011-03-07 06:34:14Z

0

What are you adding to your array?

If you are setting you php like that :

$headincludes=array("$var1","$var2");

it will not work. You have to do like that:

$headincludes=array($var1,$var2);

answered Mar 7, 2011 at 6:34

j_freyre

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1 Comment

alex Over a year ago

"$var1" actually should work, even if interpolating it in a string is unnecessary.

deceze · Accepted Answer · 2011-03-07 06:34:41Z

0

Execute the PHP code in your main files and dump the result into $headincludes. There's no need to execute the code in the header, the result won't change.

Or would it? If you're actually trying to inject arbitrary code into other code, you're doing it wrong.

answered Mar 7, 2011 at 6:34

deceze♦

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Comments

xkeshav · Accepted Answer · 2011-03-07 06:35:10Z

0

whats into $headincludes

note: use include_once

answered Mar 7, 2011 at 6:35

xkeshav

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Comments

yossi · Accepted Answer · 2011-03-07 06:35:39Z

0

did you try string concatenate. like this

$headincludes=array("<your html code>".$variable."<more html>","<more>")

answered Mar 7, 2011 at 6:35

yossi

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Comments

grc · Accepted Answer · 2011-03-07 06:39:07Z

0

Try this:

echo eval('?>html or javascript<?');
echo eval('php');

answered Mar 7, 2011 at 6:39

grc

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2 Comments

deceze Over a year ago

Because echo eval('?><div>foobar</div><?') is so much easier than echo '<div>foobar</div>' or ?><div>foobar</div><??

grc Over a year ago

Because it allows either php or html to be in the array $headincludes=array("$x = 2+3;","?><div>foobar</div><?");

Alexander · Accepted Answer · 2012-03-07 08:31:27Z

0

If you try to output php file, sure you do that easily but your browser won't able to run.

So you need to:

Eval the php file and output the result

or

Create somehow html file that you'll output instead of php file.

include_once() or include() are your friends anyway, it'll (just like name reads) include the file and the file is "executed" in that very place. "once" version won't include the same file twice if it was already included before (useful if you do several includes, which in turn do another inclusions).

answered Mar 7, 2012 at 8:31

Alexander

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