I am confused about default values for PHP functions. Say I have a function like this:
function foo($blah, $x = "some value", $y = "some other value") {
// code here!
}
What if I want to use the default argument for $x and set a different argument for $y?
I have been experimenting with different ways and I am just getting more confused. For example, I tried these two:
foo("blah", null, "test");
foo("blah", "", "test");
But both of those do not result in a proper default argument for $x. I have also tried to set it by variable name.
foo("blah", $x, $y = "test");
I fully expected something like this to work. But it doesn’t work as I expected at all. It seems like no matter what I do, I am going to have to end up typing in the default arguments anyway, every time I invoke the function. And I must be missing something obvious.
I would propose changing the function declaration as follows so you can do what you want:
This way, you can make a call like
foo('blah', null, 'non-default y value');and have it work as you want, where the second parameter$xstill gets its default value.With this method, passing a null value means you want the default value for one parameter when you want to override the default value for a parameter that comes after it.
As stated in other answers,
If I have a method that can accept varying numbers of parameters, and parameters of varying types, I often declare the function similar to the answer shown by Ryan P.
Here is another example (this doesn’t answer your question, but is hopefully informative: