Notice: Undefined variable: var1 in \www\test.php on line 7

Question

Possible Duplicate:
How to use include within a function?

i have two file

inc.php file

<?php
$var1 = "foo";
$var1 .= "bar";
?>

test.php File

<?php
function getcontent($file) {
 include ($file);
}

getcontent('inc.php');
echo $var1;
?>

When i run test.php it give me error in output

Notice: Undefined variable: var1 in \www\test.php on line 7

But when i change my test.php file to this

<?php
include ('inc.php');
echo $var1;
?>

its works, and give me output fine

foobar

score 1 · Accepted Answer · edited May 23 '17 at 11:56

1

When you do it by this

function getcontent($file) {
    include ($file);
}
getcontent('inc.php');

it includes as

function getcontent($file) {
    $var1 = "foo";
    $var1 .= "bar";
}

actually your variables are being included inside the function and not visible outside of the function so the error message occurs.

edited May 23 '17 at 11:56

Community

answered Jun 24 '12 at 17:19

The Alpha

so this will not work i include in inc.php??? – user1452926 Jun 24 '12 at 17:27
Not outside the function if it's not a global variable. – The Alpha Jun 24 '12 at 17:28

Marcio Mazzucato · Answer 2 · 2012-06-24T17:24:40.313

0

You have to declare $var1 as global var.

global $var1;

$var1 = "foo";
$var1 .= "bar";

OR

$GLOBALS['var1'] = "foo";
$GLOBALS['var1'] .= "bar";

edited Jun 24 '12 at 17:24

answered Jun 24 '12 at 17:14

Marcio Mazzucato

He doesn't have $var2 ;-). I'm sure you have ment $var1 only. – Stan Jun 24 '12 at 17:15
@Stan, Thanks to observe, i've done an update =) – Marcio Mazzucato Jun 24 '12 at 17:20

score 0 · Answer 3 · answered Jun 24 '12 at 17:15

0

when you include the file in the getcontent function, the vars behave like they were defined there and aren't visible to the outside.

if you declared them as global, it would work.

answered Jun 24 '12 at 17:15

Stefan S

3 Answers3