Undefined variable?

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

Set the $abc into the function

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

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

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
$abc = 'hello';

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

echo xyz($abc);

?>


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

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

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();
?>

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