Need a little help
I have
$_POST["zapremina"]=2000;
$_POST["starost"]="15%";
$_POST["namena"]="50%";
I want simple function to do this
$foo=(2000 - 15%) - 50%;
How to do that?
4 Answers
PHP is loosely typed, so you don't have to cast types explicity or do unnecessary operations (e.g. str_replace)
You can do the following:
$z = $_POST["zapremina"]; //$_POST["zapremina"]=2000;
$s = $_POST["starost"]; //$_POST["starost"]=15%;
$n = $_POST["namena"]; //$_POST["namena"]="50%;
$result = (($z - ($z *($s / 100))) - ($z * ($n / 100)));
Remember to use parentheses to have a readable code and meaningful var names.
Comments
Like this:
$starostPercentage = (substr($POST["starost"], 0, -1) / 100);
$namenaPercentage = (substr($POST["namena"], 0, -1) / 100);
$foo = ($_POST["zapremina"] * (100 - $starostPercentage)) * $namenaPercentage;
This is what this does and why:
- Convert the percentages (like
15%) from their text form to their decimal form (substr(15%)=15,15 / 100=0.15). - Calculate
$foowith these decimals.2000 - 15%is what you would write (as a human), but in PHP you need to write that as2000 * (100 * 0.15), meaning: 85% of 2000).
1 Comment
Shikiryu
You don't take
$_POST["zapremina"] into account. Plus, it will result as : (2000 * 15) -50I'd go with this:
$zap = intval($_POST['zapremina']);
$sta = intval($_POST['starost']);
$nam = intval($_POST['namena']);
$foo = ($zap * ((100-$sta)/100)) * ((100 - $nam)/100)
answered Mar 30, 2013 at 19:27
Bart Friederichs
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Comments
add this function and then call it
function calculation($a, $b, $c)
{
$b = substr($b, 0, -1) / 100;
$c = substr($c, 0, -1) / 100;
return (($a * $b) * $c);
}
and now you can call
$foo = calculation($_POST["zapremina"], $_POST["starost"], $_POST["namena"]);
go with function most of times, because it will be helpful for reusability.
$_POST["zapremina"]="2000"; $_POST["starost"]='15%'; $_POST["namena"]="50%";?$a= (int) $b;