Question

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Any way to specify optional parameter values in PHP?

just randomly came across this.

If I have a function like so:

public function getSomething($orderBy='x', $direction = 'DESC', $limit=null){

//do something random

}

When calling the function is it possible to ignore the first two fields and leave them default yet specify the 3rd.

For example:

$random = $this->my_model->getSomething(USE_DEFAULT, USE_DEFAULT, 10);

I know I can pass the 1st and 2nd parameters but all im asking is if their is some kind of special keyword that just says use the default value.

Hope that makes sense. its not a problem, just curious.

thanks for reading

Answer 1

You need to do that yourself. You can use null to indicate that a default value should be used:

public function getSomething($orderBy = null, $direction = null, $limit = null) {
    // fallbacks
    if ($orderBy === null) $orderBy = 'x';
    if ($direction === null) $direction = 'DESC';

    // do something random
}

Then pass null when calling it to indicate that you want to use the defaults:

$random = $this->my_model->getSomething(null, null, 10);

---

Another possible solution that I use sometimes is an additional parameter at the very end of the parameter list, containing all optional parameters:

public function foo($options = array()) {
    // merge with defaults
    $options = array_merge(array(
        'orderBy'   => 'x',
        'direction' => 'DESC',
        'limit'     => null
    ), $options);

    // do stuff
}

That way you do not need to specify all optional arguments. array_merge() ensures that you are always dealing with a complete set of options. You would use it like this:

$random = $this->my_model->foo(array('limit' => 10));

It seems like there is no required parameter this particular case, but if you need one, simply add it in front of the optional ones:

public function foo($someRequiredParameter, $someOtherRequiredParameter, $options = array()) {
    // ...
}