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.
try this:
function test($required, $optional = NULL){..}
then you can call
test($required, $optional)
and with $optional null
test($required);
$required AND '$optional' with values, this $optional=NULL would be overwritten? will work perfectly with the $optinal data?1) You can define default values for arguments like so:
function test($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
You can also do this:
function test($required, $optional = ''){
#code
}
when calling:
test($required)
or:
test($required, $optional = 'something')
test($required, $optional = 'something') is misleading, you're assigning $optional variable, not passing optional argument value by name
function f($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 $b='b'just set the default parameters in your function ex.
function test($required, $optional = null, ... )
now you can call your function like this
test($required, $optional)
test($required);
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");
?>
