Undefined variable?

[cancer10]

HI People

How come the following code generates undefined variable error?

Error:

Notice: Undefined variable: abc in D:\xampp\htdocs\test2.php on line 6

<?php 
error_reporting(E_ALL);
 $abc  = 'hello';

function xyz(){
	echo $abc;
}

echo xyz();

?>

Any help will be appreciated


Thanks

[er4o]

Set the $abc into the function

PHP Code:

<?php 
function xyz() {
$abc  = 'hello';
return $abc;
}
echo xyz();
?>

[cancer10]

No, I want to access a variable that is declared outside a function, within that function. can we?

[thekooliest]

You have to pass variables into a function, they cannot read variables created outside. So now, function xyz reads the first parameter entered (in this case $abc is sent in the line echo xyz($abc) and uses it as variable $test within the function xyz...

PHP Code:

<?php 
$abc  = 'hello';

function xyz($test){
    echo $test;
}

echo xyz($abc);

?>

http://www.w3schools.com/php/php_functions.asp (scroll down to adding parameters)

[Dormilich]

you could also make use of the superglobals (in this case the $GLOBALS array) that PHP provides.

[dmgroom]

For a start you effectively are duplicating the echo command, once within the fuction and once when calling the function. Secondly you need to declare the variable global within the function.

<?php 
error_reporting(E_ALL);
$abc  = 'hello';

function xyz(){
   global $abc;
    echo $abc;
}

xyz();
?>

OR

<?php 
error_reporting(E_ALL);
$abc  = 'hello';

function xyz(){
   
    echo $GLOBALS['abc'];
}

 xyz();
?>

[Fou-Lu]

globals destroy reusability. Pass you're variable as a parameter to you're function instead.