Letting user input PHP code and having a function parameter act on it

[BenGoldberg]

One more question. So I have a function with two parameters, $x and $y. Here it is.

 

function dydx($x,$y){

$equation = 2 * $x;
return $equation;
}

 

Now here's the problem. I want $equation to be user defined. Easy enough, I use a post command and I get $equation to equal whatever the user inputs. The problem is that if I get input from the user, I'm not sure how to take that input and then have the function parameters work on it. Like if the user inputs "3*$x + 2*$y", i want to be able to let the parameters for the function dydx act on it. How could I go about doing this?

[anupamsaha]

Read the eval() in PHP http://php.net/manual/en/function.eval.php. But, be very careful because anybody can input any PHP command to blow your server. Sanitize users' input before you use them.

[BenGoldberg]

Eval seems to be the right function for my application, but for the life of me I can not get it to work. Here's the basic code I'm working with.

 

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">

<table>
<tr>
	<td>Enter function y prime:</td>
	<td><input type="text" size="50" maxlength="" name="yprime" value="<?php echo (isset($_POST['yprime']) ? $_POST['yprime'] : '')?>" /></td>
</tr>

</table>

</br>

<input type="submit" name="rk" value="Submit" />

</form>

<?php

$yprime = strval( $_POST['yprime'] ); //saving the function as a string

function dydx($x,$y){

eval("\$equation = \"$yprime\";");
return $equation;
}

echo dydx(3,2); //equals 0 when it should equal 9 (yprime being 3*$x)

?>

 

(The example input i use for yprime is 3*$x)

 

I've done hours of googling and researching how eval works, but no matter what I try, the function dydx() always returns 0. The code above is what I believe to be my best attempt at getting it to evaluate $yprime properly, but it doesn't work. Argh!

[xyph]

It's dangerous, but

 

<?php 

$eq = '3*$x + 2*$y';

$y = 3;
$x = 2;

echo eval( "return $eq;" );


?>