Re: How to assign eval() to variable?

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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@xxxxxxxx> 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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