note 21121 deleted from function.eval by didou

Note Submitter: spam@krubner.com ---- To everyone who didn't know this, and as I just learned the hard way, eval() returns nothing, so you can't do this: $str = "mysql_query('SELECT * FROM myDb')"; $databaseResult = eval ("\$str = \"$str\";"); A note above suggests using the return() function, and though I read that comment beforehand, I did not understand it. Most of us who've worked with PHP for awhile will fall into the assumption that a PHP function will return its result. Such an assumption only brings grief when you have to deal with eval(). However, I also learned that the PHP interpreter will sometimes read a string as if it was code, for reasons I don't entirely understand. I wanted a function that would walk through a 2 dimensional array and apply a PHP function to every element. I came up with the following function. Since I was passing the name of the function as a string, I at first assumed that I needed eval() to make PHP interpret the string as a function. But in the end, I found this was not so. The following function works, despite being passed a function name in a string. function processArray($dbArray, $function){ $processedArray = array(); for ($row = 0; $row < count($dbArray); $row++) { for ($element = 0; isset($dbArray[$row][$element]); $element++) { $processedArray[$row][$element] = $function($dbArray[$row][$element]); } } return $processedArray; } $db = array('dog', 'cat', 'snake'); $db2 = array('nasty', 'rotten', 'no good'); $db3 = array($db, $db2); $db3 = processArray($db3, "strtoupper");