Can we pass an array as parameter in any function in PHP?

Question

I have a function to send mail to users and I want to pass one of its parameter as an array of ids.

Is this possible to do? If yes, how can it be done?

Suppose we have a function as:

function sendemail($id, $userid) {

}

In the example, $id should be an array.

Solution 1

You can pass an array as an argument. It is copied by value (or COW'd, which essentially means the same to you), so you can array_pop() (and similar) all you like on it and won't affect anything outside.

function sendemail($id, $userid){
    // ...
}

sendemail(array('a', 'b', 'c'), 10);

You can in fact only accept an array there by placing its type in the function's argument signature...

function sendemail(array $id, $userid){
    // ...
}

You can also call the function with its arguments as an array...

call_user_func_array('sendemail', array('argument1', 'argument2'));

Solution 2

even more cool, you can pass a variable count of parameters to a function like this:

function sendmail(...$users){
   foreach($users as $user){

   }
}

sendmail('user1','user2','user3');

Solution 3

Yes, you can do that.

function sendemail($id_list,$userid){
    foreach($id_list as $id) {
        printf("$id\n"); // Will run twice, once outputting id1, then id2
    }
}

$idl = Array("id1", "id2");
$uid = "userID";
sendemail($idl, $uid);

Solution 4

Yes, you can safely pass an array as a parameter.

Solution 5

What should be clarified here.

Just pass the array when you call this function.

function sendemail($id,$userid){
Some Process....
}
$id=array(1,2);
sendmail($id,$userid);

View more solutions

Comments

[Gaurav]

@vickirk : I edited your answer to correct mistake of $userid. hope you don't mind?

[Tanner Ottinger]

Note that you don't need to define it before calling the function. you could have done: sendemail(array("foo" => "bar"), ...).

[Gaurav]

@Anonymous Loozah : when length of array is not too big.

[Tanner Ottinger]

@Gaurav : Yeah, but the array is probably going to be generated anyway.

[vickirk]

Duh, I had just been checking £/$ rates, maybe my fingers got confused when my brain was asleep. Cheers @Gaurav

[shareef]

This better solution works with type safe e.g (Event ...$event)