Passing an optional parameter in PHP Function

Question

How can I pass an optional parameter to a PHP function. For example

function test($required, $optional){..}

So that i should be able to call the function by passing either one parameter or both. For example:

test($required, $optional)
test($required);

Thanks.

wow almost every single one of the answers for this question say the exact same thing. I don't know how you managed to pick one because they're all correct. :) — edwardtyl, Jul 22 '15 at 18:51

score 55 · Accepted Answer · answered Mar 26 '11 at 11:10

55

try this:

function test($required, $optional = NULL){..}

then you can call

test($required, $optional)

and with $optional null

test($required);

answered Mar 26 '11 at 11:10

felixsigl

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You mean when I pass `$required` AND '$optional' with values, this `$optional=NULL` would be overwritten? will work perfectly with the `$optinal` data? – Neocortex Dec 16 '14 at 06:32

score 11 · Answer 2 · answered Mar 26 '11 at 11:10

11

With a default value:

function test($required, $optional="default value"){..}

answered Mar 26 '11 at 11:10

Skilldrick

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score 8 · Answer 3 · answered Mar 26 '11 at 11:12

8

1) You can define default values for arguments like so:

function test($required, $optional = false) {}

2) You can use func_num_args() to get the number of arguments, func_get_arg($n) to get the n'th argument or func_get_args() to get an array of all arguments.

You can find a good summary here: Function arguments

answered Mar 26 '11 at 11:12

Alp

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+1 for mentioning another way – Your Common Sense Mar 26 '11 at 11:16
yes +1 for mentioning another way. It sure beats a repeat answer for an entire page. – Stephen Nielsen Jun 17 '14 at 16:54

score 6 · Answer 4 · answered Feb 13 '14 at 14:47

6

You can also do this:

function test($required, $optional = ''){
    #code
}

when calling:

test($required)

or:

test($required, $optional = 'something')

answered Feb 13 '14 at 14:47

Limon

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calling with `test($required, $optional = 'something')` is misleading, you're assigning `$optional` variable, not passing optional argument value by name – McX Mar 30 '18 at 09:50
@McX does it work? – Limon Mar 30 '18 at 15:03
1

@Lemon It works with a side effect which if not well understood can lead to errors. For example someone may try to call `function f($a='', $b='')` with `f($b='b')` and think it will call `f('', 'b')` when it will actually call `f('b', '')` and set the global variable `$b='b'` – McX Apr 02 '18 at 07:22
@McX that's a good point, but I guess it has more to do with being able to pay attention to the function you want to call. That can be solved by just adding 1 line of comment, preventing this unwanted side effect. – Limon Apr 03 '18 at 12:00

score 2 · Answer 5 · answered Mar 26 '11 at 11:10

2

just set the default parameters in your function ex.

function test($required, $optional = null, ... )

now you can call your function like this

test($required, $optional)
test($required);

answered Mar 26 '11 at 11:10

mjspier

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score 0 · Answer 6 · answered Mar 26 '11 at 11:12

read this http://php.net/manual/en/function.func-get-args.php

dont pass any argument when decalare a function like below

function myMin() {
      $list = func_get_args();
      $min = NULL;
      if (count($list)>0) $min = $list[0];
      for ($i=1; $i<count($list); $i++) {
         if ($min > $list[$i]) $min = $list[$i];
      }
      return $min;
   }

   print("\n Calling a function that uses func_get_args():\n");
   print("    ".myMin()."\n");
   print("    ".myMin(5)."\n");
   print("    ".myMin(5, 6, 2)."\n");
   print("    ".myMin(5, 6, 2, 9, 5, 1, 4, 1, 3)."\n");
?>

Passing an optional parameter in PHP Function

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