Re: How to assign eval() to variable?

short hack, assuming your eval echo's out to the browser (which I can't see how else you'd expect something to 'return' from an eval'd statement ob_start(); eval($foo); $result = ob_get_clean(); On May 23, 2009 1:46am, Afan Pasalic <afan@afan.net> wrote:

hi,

I have on one website boxes with information, pulled from mysql. the content can be string, php code, url of other website or url to specific file etc.

currently, I have something like this:

// connect to db

// mysql_query() to get box content and content_type

switch($content_type)

{

case 'string':

echo $content;

break;

case 'php_code':

eval($content);

break;

case 'website':

echo ''.$content.';

break;

case 'file'

require_once($file);

echo $file_content;

// etc.

}

but, now I have to change the code to assign content to variable and the variable will be printed later. I tried something like this:

switch($content_type)

{

case 'string':

$record = $content;

break;

case 'php_code':

$record = eval($content);

break;

case 'website':

$record = ''.$content.';

break;

case 'file'

require_once($file);

$record = $file_content;

// etc.

}

and it works - except eval() part. cant do $record = eval($content); ?!?!?!?

thanks

afan

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