php basic calculator with function and switch but result is - php

Question

i am currently doing some simple calculator for a practice but it the output or result is not showing here is my code guys hope you can help me :/
<input type="radio" value= "Addition" name="calcu"> Addition .<br />
<input type="radio" value= "Subtraction" name="calcu"> Subtraction .<br />
<input type="radio" value= "Multiplication" name="calcu"> Multiplication .<br />
<input type="radio" value= "Division" name="calcu"> Division .<br />
<?php
$num1 = $_POST['num1'];
$num2 = $_POST['num2'];
$calcu = $_POST['calcu'];
function calculate($n1,$n2)
{
switch('$calcu')
{
case "Addition";
$compute = $n1 + $n2;
break;
case "Subtraction";
$compute = $n1 - $n2;
break;
case "Multiplication";
$compute = $n1 * $n2;
break;
case "Division";
$compute = $n1 / $n2;
break;
}
}
echo "$calcu <br /> <br /> 1st Number: $num1 <br /> 2nd Number: $num2 <br /><br />";
echo "Answer is:" .calculate($num1,$num2);
?>

Answer 1

Here is complete code:
<?php
$num1 = $_POST['num1'];
$num2 = $_POST['num2'];
$calcu = $_POST['calcu'];
function calculate($n1,$n2, $calcu) // set $calcu as parameter
{
switch($calcu)
{
case "Addition": // here you have to use colons not semi-colons
$compute = $n1 + $n2;
break;
case "Subtraction":
$compute = $n1 - $n2;
break;
case "Multiplication":
$compute = $n1 * $n2;
break;
case "Division":
$compute = $n1 / $n2;
break;
}
return $compute; // returning variable
}
echo "$calcu <br /> <br /> 1st Number: $num1 <br /> 2nd Number: $num2 <br /><br />";
echo "Answer is:" .calculate($num1,$num2, $calcu); // you need to pass $calcu as argument of that function
?>

Answer 2

Change switch('$calcu') to switch($calcu). It should be this way.
But not only that. Your variables are undefined because you are trying to address them before form is submited, i.e they don't exist yet.
$num1 = $_POST['num1'];
$num2 = $_POST['num2'];
$calcu = $_POST['calcu'];
And there you address them
echo "$calcu <br /> <br /> 1st Number: $num1 <br /> 2nd Number: $num2 <br /><br />";
echo "Answer is:" .calculate($num1,$num2);
The right way to implement this is to check if form was submitted:
<input type="radio" value= "Addition" name="calcu"> Addition .<br />
<input type="radio" value= "Subtraction" name="calcu"> Subtraction .<br />
<input type="radio" value= "Multiplication" name="calcu"> Multiplication .<br />
<input type="radio" value= "Division" name="calcu"> Division .<br />
<?php
if (isset($_POST)){
$num1 = $_POST['num1'];
$num2 = $_POST['num2'];
$calcu = $_POST['calcu'];
function calculate($n1,$n2)
{
switch('$calcu')
{
case "Addition";
$compute = $n1 + $n2;
break;
case "Subtraction";
$compute = $n1 - $n2;
break;
case "Multiplication";
$compute = $n1 * $n2;
break;
case "Division";
$compute = $n1 / $n2;
break;
}
}
echo "$calcu <br /> <br /> 1st Number: $num1 <br /> 2nd Number: $num2 <br /><br />";
echo "Answer is:" .calculate($num1,$num2);
unset($_POST);
}
?>

Answer 3

Change switch('$calcu') to switch($calcu).
As #PeterM mentioned, you are accessing variable $calcu out of scope. Either you pass the $calcu variable to fun calculate or access directly by $_POST array.
use switch($_POST['calcu']).
OR
function calculate($n1,$n2, $calcu) {
...
}
Call the fun by calculate($n1,$n2, $calcu).