Use my php variable as a formula to calculate something

Question

I'm working in an invoice system, that has to calculate different formulas for every product, which will need to be filled with Width, Height, and Material cost.

I did a database where i'm storing

t_title -> "Curtain 1"

t_formula -> "({ANCHO}*{ALTURA})*{PRECIO}"

and then php does this:

<?php 
   $ancho = str_replace("{ANCHO}", $_POST['ancho'], $articulo['t_formula']);
   $alto = str_replace("{ALTURA}", $_POST['alto'], $ancho);
   $precio = str_replace("{PRECIO}", $_POST['precio'], $alto);
   $total = $precio; echo eval($total);
?>

and the result is not giving anything, but a blank space.

How can i make it to work? I know that php can't calculate from variables as php but, i can't find another way to do it.

`*`s are formatting elements in markdown. I edited the question so the variable is displayed properly. — Mureinik, Dec 06 '14 at 16:06
Be mindful using [`eval()`](http://stackoverflow.com/questions/951373/when-is-eval-evil-in-php) — Jason McCreary, Dec 06 '14 at 16:08

score 1 · Answer 1 · answered Dec 06 '14 at 16:14

The eval() function expects the string parameter to be a proper code statement.

$str = '(10*10) * 1000';
echo eval($str);

will give you an error, because php cannot evaluate that string. However,

$str = 'return (10*10) * 1000;';
echo eval($str);

will evaluate the expression correctly.

You should use:

$total = 'return ' . $precio . ';';
echo eval($total);

score 0 · Answer 2 · answered Dec 06 '14 at 16:12

0

Your using eval is wrong. The string passed to eval must be valid PHP code. i.e:

$precio2 = '$total = '.$precio.';';
eval($precio2);
echo $total;

http://php.net/manual/en/function.eval.php

answered Dec 06 '14 at 16:12

SaidbakR

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@davco98 take care with the edit of the answer, I added `;` at the last to the string to be valid PHP code. – SaidbakR Dec 06 '14 at 16:17

Use my php variable as a formula to calculate something

2 Answers2