TalkPHP - View Single Post - User defined function with 'return' statement -- not working

Wildhoney · 04-15-2008, 06:12 PM

The only segment that I have modified is assigned the value of the salestax function to the $nYourtotal variable. When you issue the return in your custom function, that simply returns data back, and so you can either assign (as in our case), print (as we do later on when we print), or do nothing with (as you were doing previously).

php Code:

// Charge for an item.
$nCharge = 100.00 ;

// Define the function.
function salestax($nCharge, $nShip = .10)
{
    $nYourtotal = ($nCharge + ($nCharge * $nShip));
    return $nYourtotal ;
}

// Invoke the function...
$nYourtotal = salestax($nCharge) ;

// Print the results...
print $nYourtotal ;

04-15-2008, 06:12 PM	#3 (permalink)
Wildhoney La Vida es Sueño Join Date: Sep 2007 Location: Oldham Posts: 2,280 Thanks: 90	The only segment that I have modified is assigned the value of the `salestax` function to the `$nYourtotal` variable. When you issue the `return` in your custom function, that simply returns data back, and so you can either assign (as in our case), print (as we do later on when we `print`), or do nothing with (as you were doing previously). php Code: // Charge for an item. $nCharge = 100.00 ; // Define the function. function salestax($nCharge, $nShip = .10) { $nYourtotal = ($nCharge + ($nCharge * $nShip)); return $nYourtotal ; } // Invoke the function... $nYourtotal = salestax($nCharge) ; // Print the results... print $nYourtotal ; __________________ The man who comes back through the Door in the Wall will never be quite the same as the man who went out.