Eval is messed up

CoB-Himself

I have code that goes like this:

<?php
$code = "echo 1;";

$evaluated = eval("$code");

echo $evaluated;

The output returns:

What it is doing is when I call the eval() function it automatically outputs it. I don't want it to output what it is evaluating until I call the variable that I assigned the results to ($evaluated). Does anyone have any idea why it is doing that and how I can fix it?

Thanks.
CoB

leatherback

Hi,

Looks like there is a problem with the way you are trying to fill the variable.

Have a look at the manual (http://www.phpbuilder.com/manual/function.eval.php )

and try:

eval("\$evaluated = \"$code\";");

echo $evaluated;

(Not sure about this: Never used eval myself)
J

CoB-Himself

That didn't work... it just returned $evaluated as echo 1; which is what $code is equal to.

Thanks for your help though.

Anymore comments? Anyone?

CoB

Spear_Chucka

I think leatherback is right. Check out the php site and you'll see it described like this

<?php

$num = "1";
$code = "echo $num";

echo $code;

eval("\$code = \"$code\";");

echo $code;

?>

it should return:

echo $num
echo 1

leatherback

Hi,

Maybe there are other ways of achieving what you are trying. What do you want to accomplish? (I do assume you are not really trying to echo 1 to the screen, but that this is a simplified situation of what you try to do?)

CoB-Himself

I have fields in a database that contain regular text and then some php tags with the codes. I am extracting the code from the database and evaluating it and then replacing the tags and code with the finishing product.