PhP Variables inside functions - php

Question

I'm having troubles with variables inside functions and being able to read them in a index.php. Let's say I have this function (functions.php):
function firstFunction(){
$id = '1234';
secondFunction($id);
}
function secondFunction($idnumber){
if( $idnumber = 1234 ){
$name = "James"; //should say "james".
}
and in the index.php have something like that:
include(functions.php)
<?php
firstFunction();
echo $name;
?>
does anyone know a possible way to do this?, i need to be a able to read from a variable inside a function that is in another function.
Also when calling a function inside a function allow this to read a variable from the first function.
Thanks.

Answer 1

Variables created inside a function are local to it, you can't read their contents outside of it (because all function-local variables get deleted after the function finished running).
The improper way would be to make the $name variable global in the second function.
A better way to do it would be to use function return values: secondFunction returns $name, and firstFunction returns whatever it gets from calling secondFunction($id).
Oh, and by the way, you have another logical error in your secondFunction: Your if test will not behave like you probably intended because you are using a variable value assignment operator (=) instead of comparison (==) which always succeeds returns the assigned value that evaluates to a "true" value in most (but not all) cases.

Answer 2

This is very bad practice - using globals. The answer you need is:
global $name;
You need to put it above the row:
$name = "James"; //should say "james".

Answer 3

You declare $name inside second function, so it only exists there (that is it's "Scope").
To use it's value you need the functions to return it:
function secondFunction($idnumber){
if ( $idnumber == 1234 ){
$name = "James"; //should say "james".
}
else {
$name = "";
}
return $name;
}
function firstFunction(){
$id = 1234;
return secondFunction($id);
}
and in the index.php have something like that:
<?php
include 'functions.php';
$name = firstFunction();
echo $name;
?>

Answer 4

You can't call a variable like that it is out of scope what you need to do is as follows
function firstFunction(){
$id = '1234';
return secondFunction($id);
}
function secondFunction($idnumber){
if( $idnumber == 1234 ){
return "James";
}
}
Then in your index do this
<?php
include(functions.php);
echo firstFunction();
?>
A variable that is out of scope can not be accessed for example if you defined a variable inside a if statement is would only be able to be used in that statement and it is the same with the functions and variables that are inside. Thought I needed to explain my answer more.

Answer 5

Try this...
index.php
<?php
// Includes
include("functions.php");
// Create a variable to store the output of your function
$returned_name = firstFunction();
// Echo the variable
echo "Name: " . $returned_name;
?>
functions.php
<?php
// Function 1
function firstFunction() {
$id = "1234";
return secondFunction($id);
}
// Function 2
function secondFunction($idnumber) {
if ($idnumber == "1234") {
$name = "James";
} else if ($idnumber == "5678") {
$name = "Brian";
}
return $name;
}
?>