this function does not return the result, what could I be doing wrong?
<?php
function iisset($name){
return ${$name}; // "${$name}" does not work
}
$hola = 1;
echo iisset("hello");
Note: it works correctly if it is not in a function
You have a few issues going on here. The first is an issue of scope. Nothing outside the function iisset($name) is normally visible to it. Therefore any variable defined outside iisset($name) can't been seen.
So the first step would be to make it global
by adding global ${$name}; to the first line after the function declaration. The second would be that there is no variable named "hello" outside the function. If you are trying to access the $hola then I would recommend the following:
<?php
function iisset($name){
global ${$name};
return ${$name}; // "${$name}" does not work
}
$hola = 1;
echo iisset("hola");
Related
whether to do a declare variable when a variable is in a sub-function ?
Like the example carried this:
function cobasaja(){
global $coba;
return $coba;
}
function ditampilkan(){
global $coba;
$coba = "content trying...";
return cobasaja();
}
echo "View: ".ditampilkan();
Why it can not be like this:
function cobasaja(){
global $coba;
return $coba;
}
function ditampilkan(){
//global $coba; <= not declare in viewer function
$coba = "content trying...";
return cobasaja();
}
echo "View: ".ditampilkan();
But the second experiment did not work.
Because as I recall, usually the second way can be done, but now I do it can not, is this because of its PHP version or setting in PHP.ini ?
Adding a function creates a new scope. Any variables you want to use in the function need to be either defined in that scope, brought in from the outer scope with global, or passed in as parameters. This general concept has not changed much over PHP versions as far as I know, so I don't believe your second experiment would have worked in an earlier PHP version, or could work by changing a configuration setting.
If your functions are in the same class, you can use object properties rather than global variables to achieve something like what you want.
class Example {
private $coba = '';
protected function cobasaja() {
return $this->coba;
}
public function ditampilkan() {
$this->coba = "content trying...";
return $this->cobasaja();
}
}
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 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();
Main File;
$opid=$_GET['opid'];
include("etc.php");
etc.php;
function getTierOne() { ... }
I can use $opid variable before or after function but i can't use it in function, it returns undefined.
What should i do to use it with a function in an included file?
$getTierOne = function() use ($opid) {
var_dump($opid);
};
Its because the function only has local scope. It can only see variables defined within the function itself. Any variable defined outside the function can only be imported into the function or used globally.
There are several ways to do this, one of which is the global keyword:
$someVariable = 'someValue';
function getText(){
global $someVariable;
echo $someVariable;
return;
}
getText();
However, I'd advise against this approach. What would happen if you changed $someVariable to another name? You'd have to go to each function you've imported it into and change it as well. Not very dynamic.
The other approach would be this:
$someVariable = 'someValue';
function getText($paramater1){
return $parameter1;
}
echo getText($someVariable);
This is more logical, and organised. Passing the variable as an argument to the function is way better than using the global keyword within each function.
Alternatively, POST, REQUEST, SESSION and COOKIE variables are all superglobals. This means they can be used within functions without having to implicitly import them:
// Assume the value of $_POST['someText'] is someValue
function getText(){
$someText = $_POST['someText'];
return $someText;
}
echo getText(); // Outputs someValue
function getTierOne()
{
global $opid;
//...
}
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;
?>