php - How can I miss out parameters when calling a function?

Question

I am making a PHP script.

My script has a function that looks something like this:

function myFunction($param1 = "foo", $param2 = "bar") {}

Let's say I want to specify a value for $param2, but want to skip $param1 as the default value is fine, how could I do something like this:

myFunction(param2)?

Tl;Dr: Can I skip certain parameters when calling a function?

No. Unfortunately, php doesn't allow this. If you want to specify a parameter on the right, you always have to specify the parameters on the left. — Phil Cross
– Phil Cross, Commented May 29, 2018 at 22:42
kind of a way around this is to use an array as the argument, then you can 'skip' array keys with out issue function myFunction($param1 = array()) {} myFunction(array('a'=>7)) — user9487972
– user9487972, Commented May 29, 2018 at 23:05

Shoukat Mirza · Accepted Answer · 2018-05-29 22:54:36Z

2

When you using optional parameter you will have to put it at the end. There's no way to "skip" an argument other than to specify a default like false or null.

edited May 29, 2018 at 22:54

answered May 29, 2018 at 22:43

Shoukat Mirza

82810 silver badges21 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Jowsey Over a year ago

Ok, that's unfortunate :( Thanks though!

Collectives™ on Stack Overflow

php - How can I miss out parameters when calling a function?

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related