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, $id
should be an array.
Yes, you can do that.
function sendemail($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, $uid);
even more cool, you can pass a variable count of parameters to a function like this:
function sendmail(...$users){
foreach($users as $user){
}
}
sendmail('user1','user2','user3');
What should be clarified here.
Just pass the array when you call this function.
function sendemail($id,$userid){
Some Process....
}
$id=array(1,2);
sendmail($id,$userid);
Its no different to any other variable, e.g.
function sendemail($id,$userid){
echo $arr["foo"];
}
$arr = array("foo" => "bar");
sendemail($arr, $userid);
Since PHP is dynamically weakly typed, you can pass any variable to the function and the function will try to do its best with it.
Therefore, you can indeed pass arrays as parameters.
Yes, you can safely pass an array as a parameter.
Yes, we can pass arrays to a function.
$arr = array(“a” => “first”, “b” => “second”, “c” => “third”);
function user_defined($item, $key)
{
echo $key.”-”.$item.”<br/>”;
}
array_walk($arr, ‘user_defined’);
We can find more array functions here
In php 5, you can also hint the type of the passed variable:
function sendemail(array $id, $userid){
//function body
}
See type hinting.
<?php
function takes_array($input)
{
echo "$input[0] + $input[1] = ", $input[0]+$input[1];
}
?>
function sendemail(Array $id,$userid){ // forces $id must be an array
Some Process....
}
$ids = array(121,122,123);
sendmail($ids, $userId);
I composed this code as an example. Hope the idea works!
<?php
$friends = array('Robert', 'Louis', 'Ferdinand');
function greetings($friends){
echo "Greetings, $friends <br>";
}
foreach ($friends as $friend) {
greetings($friend);
}
?>
Source: Stackoverflow.com