Passing a variable to its parent function in PHP?

Question

I have a function w/in a function, and I need the inner function to make it's variables available in a scope of parent function, e.g.:

function sayMyName(){
    getName(); // inner function generates $name value

    echo $name; // use $name
}

sayMyName();

I could easily just globalize things w/in both functions... But my situation is far more complicated and handles more variables and globalizing each one is a bit tedious.

Thanks.

PS i noticed a lot of "return" suggestions. sorry i wasnt clear , i need to return more variables.. not a simple return. thanks guys

If you need to access a variable in such a way, most likely, your design is flawed. If such a need really exist, think about returning that variable from a function or (worse solution) using by-reference parameters. — Aleks G, Apr 17 '13 at 07:42
Why are you not working object oriented with classes and such? — silkfire, Apr 17 '13 at 07:44
Fudging hell, is it really that difficult to see that this person is most likely still learning how PHP works? Stop showering him with design and OO stuff -- you're not impressing or helping anybody. — Lee White, Apr 17 '13 at 07:48
haha i know im gona get OOP comments here.. thanks for the defense Arno, but to be honest i've been coding php for over 10 years already.. and yes i do OOP as well.. unfortunately, i didnt have formal training, everything i just picked up along the way. reading tutorials etc..i've become pretty good but i do have questions from time to time. i promised myself not to answer any more WHY NOT OOP question.. unlike others using other libraries and frameworks, i actually built my own libraries of functions and classes, MY WAY.. — BrownChiLD, Apr 17 '13 at 08:27
anyway all i really needed to know is if this was possible by some way , technique or work around or even in the new versions of PHP (ive not been updated w/ the latest) .. that's all. if not then i have tons of other ways to achieve what i want.. and yes I do use OOP, but when i need to.. like for my CRUD requirements .. but i personally dont believe / agree w/ the 100% OOP concept , specially for lots of simple stuffs i dont find .. its d same argument as WHY PUT CSS STYLING in your DIV element directly, when you can separate that in CSS block/file. — BrownChiLD, Apr 17 '13 at 08:30

BlitZ · Accepted Answer · 2013-04-17T08:26:25.870

You may use $_GLOBALS, but it`s a "bad practice". So,

1: Use return:

<?php
function getName(){
    $name = 'Smith';

    return $name;
}

function sayMyName(){
    $name = getName();

    echo $name;
}

sayMyName();
?>

Shows:

Smith

2: Use references:

<?php
function getName(&$name){
    $name = 'Smith';
}

function sayMyName(){
    getName($name);

    echo $name;
}

sayMyName();
?>

Shows:

Smith

3: Return array for multiple variables:

<?php
function getName(){
    $surname = 'Smith';
    $name    = 'John';

    return array($surname, $name);
}

function sayMyName(){
    list($surname, $name) = getName();

    echo $name, ' ', $surname;
}

sayMyName();
?>

Shows:

John Smith

4. Return custom object for multiple variables:

<?php
function getName(){
    $surname = 'Smith';
    $name    = 'John';

    $buffer = new stdClass();

    $buffer->name    = $name;
    $buffer->surname = $surname;

    return $buffer;
}

function sayMyName(){
    $obj = getName();

    echo $obj->name, ' ', $obj->surname;
}

sayMyName();
?>

Shows:

John Smith

5. Use anonymous function with use statement and references:

<?php
function sayMyName(){
    $surname = $name = 'Unknown';

    $temp = function() use (&$name, &$surname){
        $surname = 'Smith';
        $name    = 'John';
    };

    $temp();

    echo $name, ' ', $surname;
}

sayMyName();
?>

Shows:

John Smith

Wow thanks man.. bunch of ideas there .. i think ill take the returning array route and just parse that in the parent function — BrownChiLD, Apr 17 '13 at 08:33
**@BrownChiLD**, probably never thinking myself about it. :) Thanx to you. And yes, I share the concept you have written above. — BlitZ, Apr 17 '13 at 08:36

score 3 · Answer 2 · answered Apr 17 '13 at 07:42

3

do this

function sayMyName(){
    $name = getName(); // inner function generates $name value

    echo $name; // results will be returned
}

sayMyName();

I hope your inner function is returning name like this

function getName(){
    return $name;
}

then it will work

answered Apr 17 '13 at 07:42

chandresh_cool

11,753
3
30
45

thanks man, that works but i needed to send multiple vars back to the parent function, not just 1.. unless return $name,$age,$etc can work lolz. – BrownChiLD Apr 17 '13 at 08:31
1

In that case create an array and put key value pairs. Return the array and then access those values using the returned array. – chandresh_cool Apr 17 '13 at 08:32

Vyktor · Answer 3 · 2013-04-17T07:56:39.727

This is what the object oriented programming was designed for. If many functions should share variables, it is probably best to encapsulate them to class like this:

class WhateverDescibestYourViewOfTheWorld {
    protected $name;

    function __construct( $name)
    {
        $this->name = $name;
    }

    function GetName() {
        return $this->name;
    }

    function SayName()
    {
        echo $this->name;
    }
}

// And use it:
$o = new WhateverDescibestYourViewOfTheWorld();
...
$o->SayName();

Or you can build class which will be just used as data container:

class DataContainer {
    public $name;
    public $address;
    // ...
}

// By reference this will modify object previously created
function GetProperties( &$dataContainer) // Note that & isn't necessary since PHP5
{
    $dataContainer->name = "...";
}

$c = new DataContainer();
GetProperties($c);

Or you can simplify this and use array:

function GetProperties( array &$dataContainer)
{
     $dataContainer['name'] = '...';
}

$data = array();
GetProperties($data);

@ArnoSluismans yeah, you're right... I was too offensive, I've added more useful notes and examples on how to do that (still the OOP is the best solution except very rare cases). — Vyktor, Apr 17 '13 at 07:57
tbh OOP in php is very rarely the best solution but each to their own ;) — Dave, Apr 17 '13 at 08:05

Robert · Answer 4 · 2013-04-17T07:50:28.800

You can use references

    $param = "aaa";
    function f(&$param)
    {
      //dostuff
      inner($param);
      echo $param;
    }
    function inner(&$inner) { //do other stuff }

or use return value

 function f() { echo inner(); } 

 function inner($param) {return $param;}

if you work on references, both functions will work on same variable, not on a copy

http://php.net/manual/en/language.references.php

the best way would be with class

<?php 
class Person
{
 private $name;

 public function setName($name){ $this->name = $name;}
 public function sayName() {echo $this->name;}
}

$person = new Person();
$person->setName("Robert");
$person->sayName();

It's good way to make it in OOP.

score 1 · Answer 5 · answered Apr 17 '13 at 07:44

1

What about first assigning the return value of getName() to a variable?

$name = getName();

answered Apr 17 '13 at 07:44

bwoebi

23,637
5
58
79

score 1 · Answer 6 · answered Apr 17 '13 at 07:44

If you only need one variable you can do this

function getName(){
    // Some code
    return 'my name is XXX';
}
function sayMyName(){
     $name = getName(); // inner function generates $name value
     echo $name; // results to undefined
 }
 sayMyName();

Otherwise you may consider using a class : http://www.php.net/manual/en/language.oop5.php

score 1 · Answer 7 · answered Apr 17 '13 at 08:14

That what you are thinking is wrong, however you can return an array of values. For ex:

function sayMyName()
{
    $result = getName(); // inner function creates an array

    echo $result['name']; 
}

or better an object:

class Results
{
    public $name;
}

function sayMyName()
{
    $result = getName(); // inner function creating an object

    echo $result->name;
}

score 0 · Answer 8 · answered Apr 17 '13 at 07:45

0

You can also do it as below.

$name = "";

function sayMyName(){
    getName(); // inner function generates $name value
    //set $name variable inside getName() function.
    echo $name; // results to undefined
}

sayMyName();

answered Apr 17 '13 at 07:45

Dipesh Parmar

27,090
8
61
90

score 0 · Answer 9 · answered Apr 17 '13 at 07:46

0

Please use bellow code ,it will solve your problem

global $var;

You can use it anywhere within your php span.

answered Apr 17 '13 at 07:46

Atanu

77
6

...and this will also unnecessary increase the number of global scoped variables -- which is why this is not a generally recommended practice. https://stackoverflow.com/q/12445972/2943403 – mickmackusa May 06 '20 at 13:09

Passing a variable to its parent function in PHP?

9 Answers9

Linked