[RESOLVED] displaying php stored in a database

Desbrina

I have template html/php stored in a database.

I can get it to retrieve the code and the html displays fine, but no php code does. If i view source i can see the php and its in light pink

This is how it appears

<body>
<?php include'function.php'; ?>
<p>

This is how i retrieve it

<?php
	$con = mysql_connect('localhost', '*****', '*****') or die('Could not connect to database!');
    mysql_select_db('*****', $con);
	$sql = "SELECT code
	FROM user_templates
	WHERE user = '".mysql_real_escape_string($_GET['user'])."'
	LIMIT 1";
	$result = mysql_query($sql);
	if(!$result) { die("Error Loading Template Data"); }
	while ($row = mysql_fetch_array($result)) {
		print $row['code'];
	}
?>

Whys it do this?

dougal85

eval(); ?

do you want to RUN it or display the source code?

bradgrafelman

Why? Because as the manual says, [man]print/man simply outputs a string of data.

What you're trying to do is actually evaluate a string of data for PHP code, not just output it. As such, you'd want to use a function such as [man]eval/man.

Note that a common saying is "if eval() is the answer, you must have asked the wrong question", though I think it might be applicable in this situation. If others want to share their opinions/possible alternatives, I'd like to see them.

Desbrina

i tried eval and it error, and i believe its because i have html code and php code in the database

dougal85

Read the description here: http://uk.php.net/eval

"There are some factors to keep in mind when using eval(). Remember that the string passed must be valid PHP code, including things like terminating statements with a semicolon so the parser doesn't die on the line after the eval(), and properly escaping things in code_str . To mix HTML output and PHP code you can use a closing PHP tag to leave PHP mode."

Desbrina

ok, i dont really understand. Does that mean i only use an opening php tag, or leave them out completely?

dougal85

as far as I am aware. It's been a while since i played withe eval, so brad can confirm this or correct me...

You can't have either <?php or ?> in eval.

it just needs to be the code that would go between them. Since you can't have ?> you cant jump out of PHP and do HTML and so on.

I like to think of eval code as the contents of a function. if you write return $var then that will be returned where you call the eval.

ie.

$x = eval("echo 'something'; return true");

$x = true;

Desbrina

got it working, after reading the comments on the eval page, someone suggested to make it like this

print eval("?>".$row['code']);

and it worked

Thanks to those that helped me

bradgrafelman

Don't forget to mark this thread resolved.