访问除返回值以外的函数中的变量

Question

I need an help to understand this problem I am facing...and apologies if it seems silly.

I wrote a function that makes some calculations according to two differents dates (arrival and departure). It all works fine, the returned values is given and it is correct.

Nevertheless while this calculation is made there is variable created ($days) which I would like to use (echo) outside the function. I have tried to make it global but I get an error...guessing that I am on the wrong path!

So my question is how do you get a value inside a function other than the returned value? if it is at all possible of course.

code below:

function costs($date1, $date2, $price) {

$arr= explode("/", $date1);
$timestamp1 = mktime(0,0,0,$arr[1],$arr[0],$arr[2]);

$arr2= explode("/", $date2);
$timestamp2 = mktime(0,0,0,$arr2[1],$arr2[0],$arr2[2]);

$timestamp = $timestamp2 - $timestamp1;
$days = $timestamp/86400;

$cost = $days * $price;

return $cost;

}     

Appreciated any little help to understand this. Francesco

Answer 1

These should really be two separate functions, with the one using the other.

function costs($date1, $date2, $price) {
    $cost = days($date1, $date2) * $price;
    return $cost; 
}

function days($date1, $date2) {
    $arr= explode("/", $date1); 
    $timestamp1 = mktime(0,0,0,$arr[1],$arr[0],$arr[2]); 

    $arr2= explode("/", $date2); 
    $timestamp2 = mktime(0,0,0,$arr2[1],$arr2[0],$arr2[2]); 

    $timestamp = $timestamp2 - $timestamp1; 

    return $timestamp/86400;
}

EDIT: Then you can call either function depending on what you want without duplicating code.

It doesn't look like you're using objects, so I won't give you the OO spiel, but you can definitely split these up into separate functions so you're not doing wierd stuff with globals and muddying up your namespaces.