PHP functions

jkobrien · 01-25-2007, 07:42 AM

Hi,

If I have a PHP function with two arguments, both with defaults, how can I just pass the second argument and rely on the default value for the first?

Code:

function test($a=1,$b=2) {
  echo "$a $b  <br />";
}

test(12);      // prints 12 2 - as you'd expect

test(,5);      // causes a blank page

test(Null,10); // prints 10
test("",10);   // prints 10

Or is this just, for some reason, a silly thing to try to do? It isn't for any particular application. I'm just learning PHP and want to understand it properly.

Many thanks in advance,

John

AdaHacker · 01-25-2007, 05:04 PM

Can't do it. Parameter passing in PHP is purely positional and you can only omit parameters at the end of the list. If you want $a to have the default value of 1 in your example, you need to use test(1, 12).

There's nothing inherently silly about wanting to do this. Some languages do allow you to omit parameters at the beginning or middle of the list. Other allow you to pass parameters by name rather than by position, e.g. test(b=12). PHP, however, is not one of those languages.

jkobrien · 01-26-2007, 04:36 AM

Thanks, AdaHacker! I like to understand these things properly.