Compare Strings given in $_POST with php

Question

I have a form that is sending in sizes of things, and I need to see what the strings are equal to so that I can set the price accordingly. When i try to do this, it says that they are not equal, and i get no prices. This is the code i'm using:

if ($_POST['sizes'] == "Small ($30)"){$total = "30";}
if ($_POST['sizes'] == "Medium ($40)"){$total = "40";}
if ($_POST['sizes'] == "Large ($50)"){$total = "50";}
else {$total = $_POST['price'];}

What am i doing wrong here? I can echo $_POST['sizes'] and it gives me exactly one of those things.

Answer 1

Try using single quotes

if ($_POST['sizes'] == 'Small ($30)'){$total = "30";}
elseif ($_POST['sizes'] == 'Medium ($40)'){$total = "40";}
elseif ($_POST['sizes'] == 'Large ($50)'){$total = "50";}
else {$total = $_POST['price'];}

Double quoted strings use variable interpolation, so the $ symbol becomes significant! See this manual page for the differences in how you can declare string literals in PHP.

(Edited to correct the logic error - as others noted, a switch would be much clearer here)

Answer 2

What Paul Dixon said is correct. Might I also recommend using a switch statement instead of that clunky chunk of if statements (which actually has a logic bug in it, I might add - $total will always equal $_POST['price'] when not 'Large ($50)')

<?php

switch ( $_POST['sizes'] )
{
    case 'Small ($30)' :
     $total = 30;
     break;
    case 'Medium ($40)' :
     $total = 40;
     break;
    case 'Large ($50)' :
     $total = 50;
     break;
    default:
     $total = $_POST['price'];
     break;
}

?>

Answer 3

That's a good candidate for a switch/case statement, with your 'else' being a default.

Also, without using elseif's on Medium and Large, if your $_POST['sizes'] is not Large, then your $total will always be $_POST['price']. This could be throwing you off as well.

Answer 4

So you know, the problem with your if/else's is that the last else is always happening. A switch is still better, but here is what your code should be:

if ($_POST['sizes'] == "Small ($30)") { $total = "30";
} else if ($_POST['sizes'] == "Medium ($40)") { $total = "40";
} else if ($_POST['sizes'] == "Large ($50)") { $total = "50";
} else { $total = $_POST['price']; }

To everyone that says the problem is the $30, $40, etc, it's not. Variables can't start with a number so PHP will ignore the $40, etc.

Answer 5

Is $total a string?

$total = "30"; is the syntax for a string. $total = 30; would be correct for Numeric.

Answer 6

Isn't there a security hole here? What if someone just submits whatever price they want for the default clause?

Answer 7

Or, even better than the clunky switch, you can take advantage of this simple logic and practise 'data-driven programming':

$vals = array(
    'Small ($30)' => 30,
    'Medium ($40)' => 40,
    'Large ($50)' => 50
);

$total = array_key_exists($_POST['sizes'], $vals)
    ? $vals[$_POST['sizes']]
    : $_POST['price'];

Answer 8

// remove any non-decimal characters from the front, then extract your value,
// then remove any trailing characters and cast to an integer
$total = (integer)preg_replace("/^\D*(\d+)\D.*/", "$1", $_POST['sizes']);
if (!$total) $total = $_POST['price'];

Answer 9

Apart from the actual cause of this error, it could have been avoided if you had used other values than the labels, e.g.:

<select name="sizes">
    <option value="small">Small ($30)</option>
    <option value="meduim">Medium ($40)</option>
    <option value="large">Large ($50)</option>
</select>