Access variable inside function from another function ?!

[vivis933]

It is possible to make it like the example below ?

function funct1($number1, $number2)
{
$variable1 = $number1 + $number2 ;
}


function funct2($variable1 , $number3)
{
$variable2 = $variable1 / $number3 ;
}

[requinix]

The code is syntactically valid.

 

If I knew what you were trying to do with it, I would make a comment pertaining to that.

[vivis933]

This an example of what i need , basically the variable from first function to be included in the second function.

<?php
$number1 = 6 ;
$number2 = 4 ;
$number3 = 2 ;
funct1($number1, $number2);
funct2($variable1 , $number3);
echo $variable2;

function funct1($number1, $number2)
{
$variable1 = $number1 + $number2 ;
}


function funct2($variable1 , $number3)
{
$variable2 = $variable1 / $number3 ;
}
?>

ERROR 

 

PHP Notice:  Undefined variable: variable1 on line 6

PHP Notice:  Undefined variable: variable2 on line 7

 

Edited September 9, 2016 by vivis933

[Jacques1]

Well, this is not how variables work. A local variable is only visible within that function and gets destroyed immediately when the function returns.

 

I think what you actually want to do is return the result of the addition so that you can then use it outside of the function:

<?php

function add($number_1, $number_2)
{
    return $number_1 + $number_2;    // add the two numbers and return the result
}

function div($number_1, $number_2)
{
    return $number_1 / $number_2;
}



$sum = add(6, 4);
$quot = div($sum, 2);

echo $quot;

[vivis933]

Thank you for the effort i will try