Change function arguments inside function

Question

HTML form sends empty values as empty strings. I want them to be null:

function dummy($arg1, $arg2, ... $argN)
{
   if(!$arg1) $arg1 = null;
   if(!$arg2) $arg2 = null;
   ... 
   if(!$argN) $argN = null

}

This is very ugly. I want something like:

function dummy($arg1, $arg2, ... $argN)
{
   nullEmptyArguments();
   var_dump($arg1); //null
}

Why set the arguments to null? If they are only set to null when they evaluate to false, then simply checking the condition (!$arg1) is equivalent to checking, setting to null, and later checking for null. — Rebecca Nelson
– Rebecca Nelson, Commented Oct 8, 2013 at 13:27
This is not a problem you typically need a solution for. Sounds like you should structure your code inside your function differently. It really hardly makes sense to do what you are doing. — deceze
– deceze ♦, Commented Oct 8, 2013 at 13:41

Glavić · Accepted Answer · 2013-10-08 13:33:06Z

1

Use function get_defined_vars() to get all defined variables, and loop over them and re-set them:

function dummy($arg1, $arg2, $argN)
{
    foreach (get_defined_vars() as $k => $v) $$k = $v ?: null;
    // your logic
}

dummy(1, '0', ''); # $arg2 & $argN will be set to NULL

answered Oct 8, 2013 at 13:33

Glavić

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4 Comments

jscripter Over a year ago

good point from the $$, because with $v doesn't work. but what about the &$ ?

Glavić Over a year ago

@BubuDaba: I don't understand what doesn't work with $v? And what do you mean by &$? What would you like to pass by reference?

jscripter Over a year ago

Oh I meant if dummy($a,$b,$c), to change the variables

Glavić Over a year ago

If you would use function dummy(&$arg1, &$arg2, &$argN) then your input must be variables like you shown, and not values, like in my answer, because you would get Fatal error: Only variables can be passed by reference.

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