[Resolved] Getting stored PHP code to display correctly

macca_25

Hi all,
I'm wondering if anyone can tell me if there is a function for the following?
I have a php page that displays HTML code stored in a mysql db.
The problem i have is when i add any php code to this HTML code and insert it into the db then query the db and display the results i get the actual php code....the code is not hidden, it actaull displays it as...for example:

<? $named = "http://localhost" ?>
  <tr> 
    <td height="25" width="50%"><div align="left"><a href="http://www.swiftflix.com.au"><img src="<? echo $named ?>/cart_uploads/images/swiftflix_logo.gif" alt="SWIFTFLIX - Australia's Premiere Online DVD Shop" width="130" height="25" border="0"></a></div></td>

The above is a snippet of the code if you look at the source in IE.

Any idea how to get the php code within the HTML as php and not just text?
Cheers,
macca

sidney

eval()

http://www.php.net/eval

macca_25

Thanks for that sidney.

If i use the eval() how do i turn the variable i select from the db (which is the HTML data) into a string. At the moment it is outputing as HTML.

Cheers,
macca

sidney

would need to see some code to say for sure

<?php
$test=0;
$name = 'coffee';
$str = 'this is html** <?php if($test==1) $string = "cup"; else $string = "cup2";  echo"This is a ".$string." with my ".
$name." in it."; ?> **this is html** <?php echo"more php"; ?>'; //data from db
$bits=explode("?>",$str);
ob_start();
foreach($bits as $bit)
	{
	$php=explode("<?php",$bit);
	echo $php[0];
	if(isset($php[1]))
		eval($php[1]);
	}
$data = ob_get_clean(); 
echo $data; 
?>

the above seperates the php from the html using <?php ?> tag
echos html then evals the php

macca_25

Thanks for you time Sidney...that explained things beautifully.

Cheers,
macca

sidney

added capturing data to var rather than outputing to browser