Nee Help In Understanding! PHP Functions

Question

I'm new to php, and cannot understand the logic below.

<?php
$user = "abc";
function test(){
$user = "def";
echo $user;
}
function test_2(){
$user = "xyz";
return $user;
}
echo $user;
echo "<br />";
test();
echo "<br />";
echo test_2();
echo "<br />";
echo $user;
?>

When "echo-ing" $user, function test_2() overwrites the value of $user and prints "xyz". But when I simply echo $user [after echo test_2() ], but it prints "abc"; I mean it should again print "xyz" as object $user stores the value "xyz". Can you please explain how does the functions mechanism works here.

possible duplicate of [Reference: What is variable scope, which variables are accessible from where and what are "undefined variable" errors?](http://stackoverflow.com/questions/16959576/reference-what-is-variable-scope-which-variables-are-accessible-from-where-and) — Lanting, Jan 20 '15 at 10:18

score 0 · Answer 1 · answered Jan 20 '15 at 10:22

0

This exercise shows that the variable $user is confined to it's scope. For instance in the scope of test() the variable $user is "def", and in the scope of test_2() the variable $user is "xyz".

answered Jan 20 '15 at 10:22

Samuel Gray

119
1
5

So if I want to overwrite the values of `$user` to xyz. how can I do? – Vaibhav Sidapara Jan 20 '15 at 10:25

score 0 · Answer 2 · answered Jan 20 '15 at 10:26

0

In very basic words - the reason is because your functions are not returning back any values. Thats why your variable $user is not changing and it is remaining with the same value as defined in the beginning. Functions are separate operations which act in their own environment"

Hope this helps

answered Jan 20 '15 at 10:26

Genislav Nastev

1

`function test_2()` is returning `$user`, is it I'm doing something wrong. – Vaibhav Sidapara Jan 20 '15 at 10:33
Function test_2() is only outputting/printing the value of the variable $user it is not returning anything to the main function. – Genislav Nastev Jan 20 '15 at 11:35

score 0 · Accepted Answer · answered Jan 20 '15 at 10:37

Php is a loosely coupled language. it means you dont need to declare datatype and as well in runtime u can change the value of the instance. the first $user will print as "abc". once u called the test() function it will assign value "def" to $user. hence u returned the $user so the test() will gives the reslt as "def". when u r calling test2() function the above steps will continue and it will return the value as "xyz".

Nee Help In Understanding! PHP Functions

3 Answers3