I am new to PHP.
I'm studying variables scopes.
A variable declared outside a function has a GLOBAL SCOPE and can only
be accessed outside a function.
A variable declared within a function has a LOCAL SCOPE and can only
be accessed within that function.
The global keyword is used to access a global variable from within a
function.
To do this, use the global keyword before the variables (inside the
function)
Normally, when a function is completed/executed, all of its variables
are deleted. However, sometimes we want a local variable NOT to be deleted. We need it for a further job.
I need to declare variable within function to be global so I can get access to it from outside the function and to be static at the same time so I can keep the value of the variable after execution of the function and use it again.
I tried
global static $x;
but it doesn't work.
I need to know if I'm thinking in wrong way case I'm new to PHP.
<?php
$x = 5;
function myTest() {
echo "x is equal to".$GLOBALS['x']."";
$GLOBALS['x']++;
}
myTest();
myText();
?>
it executes only the first myTest().
and the second one display an error
Fatal error: Uncaught Error: Call to undefined function myText()
just declare it in global scope then use $GLOBALS[] array or global keyword to use that variable in a function. And as they hold the value even after function execution you don't need static keyword as well.
study $GLOBALS, Variable scope
you can use static or global to keep the value:
function doStuff() {
$x = null;
if ($x === null) {
$x = 'changed';
echo "changed.";
}
}
doStuff();
doStuff();
the result would be: changed.changed.
if you use:
function doStuff() {
static $x = null;
if ($x === null) {
$x = 'changed';
echo "changed.";
}
}
doStuff();
doStuff();
the result would be changed. because static keeps last value even if you call function in multi times
also global have the same result because of it's definition so you can also use:
global $x;
in the function and result would be same: changed.
You have typo problem in your code (second calling of your function):
function myTest() ....
Then you called it:
myTeXt();
Related
how to access the variable X in Function Func1 not in global Scope
$X='hi';
function Func1(){
global $X;
echo $X;
}
function Func2(){
$X='hello';
Func1(); // I want to echo "hello" not "hi"
}
(First of all, good job trying to avoid using a global. They are almost never the right answer.)
Variables in PHP functions are locally scoped - they don't inherit anything from where they're called. Func1 has no idea about any variables or anything else that happens in Func2, it only knows about itself.
If you want a variable available in a function, then you need to pass it in as an argument:
function Func1($X){
echo $X;
}
function Func2(){
$X='hello';
Func1($X);
}
It would be worth reading http://php.net/manual/en/language.variables.scope.php to get a basic grounding in scope in PHP.
Give this a try
$x = 'hi';
function func1(){
echo func2();
}
function func2(){
return $x = 'hello';
}
The x variable will get overridden and the function will return the variable data.
Next, the func2 will be called in func1 and the returned value will be printed on the screen.
Just tried to minimize the number of lines.
i am new in PHP and i had a question: how can i access a variable declared in a function from global scope?
function test(){
$x = 6;
$y = 5;
return $x;
}
test();
echo $GLOBALS['y'];
I want to access variable y in global.
Thank you!
Either use $GLOBALS['y'] in the function (it's super-global, so it's available anywhere) or define global $y; in the beginning of the function.
Note: using such global variables is a bad style, better return the value (or modify it by passing it as a by-reference parameter to the function)
I'm learning PHP, and I came around to the global variable concept. I don't quite understand why this variable is getting an "undefined variable" error.
function function1() {
global $totalGeneral;
$totalGeneral = 42;
}
function function2(){
echo $totalGeneral;
}
I expected 42 to be printed out. Instead I get:
Notice: Undefined variable: totalGeneral
Reading about variable scope at the PHP manual, I thought that adding "global" was enough to make the variable global.
You forgot to include the global in your second function. Without it, it is never in scope.
Just because you use the global keyword doesn't mean the rules don't apply. Global variables are always out of scope inside of a function unless you use the global keyword (or pass it as a parameter or, the case of a closure, use the use keyword).
function function1() {
global $totalGeneral;
$totalGeneral = 42;
}
function function2(){
global $totalGeneral;
echo $totalGeneral;
}
You need to make your variable global in function2() too.
Also global directive only say to php to take variable from globals, so you need to declare your variable first, so:
$totalGeneral = 69;
function function1() {
global $totalGeneral;
$totalGeneral = 42;
}
function function2(){
global $totalGeneral;
echo $totalGeneral;
}
I have two PHP files. In the first I set a cookie based on a $_GET value, and then call a function which then sends this value on to the other file. This is some code which I'm using in join.php:
include('inc/processJoin.php');
setcookie("site_Referral", $_GET['rid'], time()+10000);
$joinProc = new processJoin();
$joinProc->grabReferral($_COOKIE["site_Referral"]);
The other file (processJoin.php) will then send this value (among others) to further files which will process and insert the data into the database.
The problem I'm having is that when the grabReferral() function in processJoin.php is called, the $referralID variable isn't being defined on a global scale - other functions in processJoin.php can't seem to access it to send to other files/processes.
I've tried this in processJoin.php:
grabReferral($rid) {
global $ref_id;
$ref_id = $rid;
}
someOtherFunction() {
sendValue($ref_id);
}
But the someOtherFunction can't seem to access or use the $ref_id value. I've also tried using define() to no avail. What am I doing wrong?
you have to define the global var in the second function as well..
// global scope
$ref_id = 1;
grabReferral($rid){
global $ref_id;
$ref_id = $rid;
}
someOtherFunction(){
global $ref_id;
sendValue($ref_id);
}
felix
personally, I would recommend the $GLOBALS super variable.
function foo(){
$GLOBALS['foobar'] = 'foobar';
}
function bar(){
echo $GLOBALS['foobar'];
}
foo();
bar();
DEMO
This is a simple and working code to initialize global variable from a function :
function doit()
{
$GLOBALS['val'] = 'bar';
}
doit();
echo $val;
Gives the output as :
bar
The following works.
<?php
foo();
bar();
function foo()
{
global $jabberwocky;
$jabberwocky="Jabberwocky<br>";
bar();
}
function bar()
{
global $jabberwocky;
echo $jabberwocky;
}
?>
to produce:
Jabberwocky
Jabberwocky
So it seems that a variable first declared as global inside a function and then initalised inside that function acquires global scope.
The global keyword lets you access a global variable, not create one. Global variables are the ones created in the outermost scope (i.e. not inside a function or class), and are not accessible inside function unless you declare them with global.
Disclaimer: none of this code was tested, but it definitely gets the point across.
Choose a name for the variable you want to be available in the global scope.
Within the function, assign a value to the name index of the $GLOBALS array.
function my_function(){
//...
$GLOBALS['myGlobalVariable'] = 42; //globalize variable
//...
}
Now when you want to access the variable from code running in the global scope, i.e. NOT within a function, you can simply use $ name to access it, without referencing the $GLOBALS array.
<?php
//<global scope>
echo $myGlobalVariable; //outputs "42"
//</global scope>
?>
To access your global variable from a non-global scope such as a function or an object, you have two options:
Access it through the appropriate index of the $GLOBALS array. Ex: $GLOBALS['myGlobalVariable'] This takes a long time to type, especially if you need to use the global variable multiple times in your non-global scope.
A more concise way is to import your global variable into the local scope by using the 'global' statement. After using this statement, you can reference the global variable as though it were a local variable. Changes you make to the variable will be reflected globally.
//<non global scopes>
function a(){
//...
global $myGlobalVariable;
echo $myGlobalVariable; // outputs "42"
//...
}
function b(){
//...
echo $GLOBALS['myGlobalVariable']; // outputs "42"
echo $myGlobalVariable; // outputs "" (nothing)
// ^also generates warning - variable not defined
//...
}
//</non global scopes>
Please use global variables in any language with caution, especially in PHP.
See the following resources for discussion of global variables:
http://chateau-logic.com/content/dangers-global-variables-revisited-because-php
http://c2.com/cgi/wiki?GlobalVariablesAreBad
The visibility of a variable
I hope that helped
<?php
$a = 1;
$b = 2;
function Sum()
{
global $a, $b;
$b = $a + $b;
}
Sum();
echo $b;
?>
I want to declare a global variable using PHP and be used inside functions.
I have tried:
$var = "something";
function foo()
{
echo $var;
}
yet I receive an error stating that the $var is undefined.
How can I solve this?
$var = "something";
function foo()
{
global $var;
echo $var;
}
use the term "global" when you need to use variables that were declared outside your function scope.
PHP variables have function scope. I.e., variables inside a function can't be accessed from outside it and global variables can't (by default) be accessed from inside functions. While using the global keyword inside functions to im-/export variables is a solution, you should not do it. Functions should be self-contained; if you need a value inside a function, pass it as a parameter, if the function needs to modify global values, return them from the function.
Example:
function foo($arg)
{
echo $arg;
}
$var = "something";
foo($var);
Please read: http://php.net/manual/en/language.variables.scope.php