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PHP PHP Functions Introducing Functions PHP Function Arguments

Posted by Gremyko Coleman
Gremyko Coleman
9,756 Points

Confused about this example

I am confused, as the argument the array is named, $user_array, why when the array is declared inside the function the array is called $name instead of being called $user_array like how it was in the argument at the beginning?

<?php function welcome_dialog($user_array) { if(is_array($user_array)) { foreach($user_array as $name) { echo "Hello, $name, how are you?</br>"; } } else { echo 'Hello friends</br>'; } }

$name = array(
    'Kcin',
    'Sharon',
    'Brother',
    'Sister'
);

welcome_dialog($name);

?>

2 Answers

David Olson
David Olson
8,690 Points
David Olson
David Olson
8,690 Points

The $name array is actually set outside of your function. It is then passed into the function by adding the line welcome_dialog($name);

The $user_array argument is just a placeholder. By passing in $name it takes the place in all locations within the function where $user_array is used.

Rafael silva
Rafael silva
23,877 Points
Rafael silva
Rafael silva
23,877 Points

I will to show for you some example, let's go , firts :

//create a function function hello($arr){ if(is_array($arr)){ //using a if statement to compare if is array the $arr foreach($arr as $name){ //for each element $arr as $name, now $name is $arr ..
echo "Hello $name , how it's going!<br >\n"; } }else{ echo 'Hello friends!'; } }

$names = array( //we have create an array 'Rafael', 'Mike', 'macedo' );

hello($names); //$names is an argument inside of hello , $names is equal to $arr and $arr is equal to $name all names are inside of an array now belong to $name , I hope help you ...