-1

I'm trying to access to variables in other files within a php function. outside of the function it brings the value that i want without problems but i don't know why if i try to access to the variable inside the php function, i can't access the value.

i have a file called name.php 

<?php 
$name1 = "aa";
$name2 = "bb"; 
?>

and in index.php i'm trying to access to the variables in name.php like this. 

<?php
include_once('name.php');
echo $name1;


name();
function name(){
echo $name1;
}

?>

as i said i can print the name1 variable outside the function but not inside. can i know why and how to access to the variables in this case?

Thank you

Masivuye Cokile
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James David Deann
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  • I had a similar issue recently and solved it by changeing the variables to globals and writing ```global $name1;``` in the first line of the function. – Jonathan Jan 16 '18 at 08:31
  • Other way, try passing it as parameter to name(); – Wocugon Jan 16 '18 at 08:33
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    Instead of using `global` (which you should avoid if you can), just pass the name to the function as an argument. – M. Eriksson Jan 16 '18 at 08:33
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    You needs to go learn some of the essential language basics ... http://php.net/manual/en/language.variables.scope.php – CBroe Jan 16 '18 at 08:33
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    Possible duplicate of [PHP access global variable in function](https://stackoverflow.com/questions/15687363/php-access-global-variable-in-function) – Darragh Enright Jan 16 '18 at 09:07

5 Answers5

2

Your problem comes from a basic of PHP : the scope of functions is local, not global, contrary to JavaScript.

You could use globals to achieve what you want to do, but I strongly discourage you to use them.

To me, your best shot is to pass your variables directly to your function, or to use closures.

// Example 1 : pass variables to the function
function name($name1) {
    echo $name1;
}

// Example 2 : use a closure
$myNameFunc = function() use($name1) {
    echo $name1;
};

You can read more about closures from the PHP documentation.

pyrsmk
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1

In PHP (and many other languages) the variables are visible only in the scope in which they are defined. In your example: 

<?php 
$name1 = "aa";
$name2 = "bb"; 
//this is the *global* scope
?>


<?php
include_once('name.php');
echo $name1; // you are still in the global scope so $name1 is defined


name();
function name(){
echo $name1; // here you are in the name() function scope. The variable $name is not defined here
}

?>

There are 2 ways to access variables from the global scope inside a function. One them is using the keyword global when defining your variable. That makes the variable visible inside all scopes which are inside the "global" scope. That is not recommended since the variable will be visible in scopes in which you don't need it (and other reasons maybe someone could explain better).

The best thing you can do is to pass your variable as a parameter to your function.

You can define the function like this:

function name($param){
    echo $param;
}

And later call your function like this:

name($name1);
VTodorov
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0

You can use Global variable or you can add parameter to your function to pass the value.

By using Global:

echo name();
function name(){
  global $name1, $name2;
  echo $name1;
}

By Adding function parameter:

echo name($name1);
function name($name1){
  echo $name1;
}
jun drie
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0

You are not passing values to the function. Try this

function name($name1){
echo $name1;
}
Maneth Pathirana
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-1

Use global $name; inside function

Read more about globals Variable.

TarangP
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Raf A.
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  • Avoid globals if you can. In this case, just passing the variable to the function as an argument is a much better way. – M. Eriksson Jan 16 '18 at 08:35