Unable To Access PHP Global Variable - php

Question

I am trying to pull global variable $nsfw but it shows nothing at all. If I echo it inside function, it works. But outside, it fails even when defined as global. Kindly help me out here.
<?php
if(!function_exists('do_example_work'))
{
function do_example_work()
{
global $nsfw;
include("includes/dbconnect.php");
// Create connection
$conn = new mysqli($DBHOST, $DBUSER, $DBPASS, $DBNAME);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT fieldname FROM table WHERE name='$anything'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
if ($row["fieldname"] == 1) {
// do somethin
$nsfw = 25;
exit();
} else {
echo "enjoy";
}
}
} else {
echo "0 results";
}
}
echo $nsfw;
};
?>

Answer 1

There are arguments for and against Globals. You can search Stack and the internet and read about:
Globals are bad in many ways, and should be avoided where possible
Globals are not bad, and can be fine/safe to use if you know what you are
doing
The manual:
http://php.net/manual/en/functions.user-defined.php
You seem to be over complicating things with this basic user defined function.
OUT
If you want to get data out of a function, just use the return statement
function do_example_work() {
// Do some stuff here
$nsfw = 25;
return $nsfw; // Return where needed, in conditional statement or end of function
}
// Will echo "25"
echo do_example_work();
IN
FYI:
To get data into the function from outside, just pass the data into your function from the outside as an argument:
function do_example_work($nsfw) {
/** The var "$nsfw" will have whatever data you pass in through the function call
* You can use it as required - check if $nsfw == something
* Or it might be database login details (urgh)
*/
echo $nsfw." - And words from in the function";
}
// Will echo "Pass in argument - And words from in the function"
do_example_work("Pass in argument");

Answer 2

The keyword global in front of a variable name means that the variable is defined somewhere outside the function, and now I want to use that variable. It does not generate a global variable. So you need to define the variable outside the function, and then you can use that variable inside a function using the global keyword in front of it.
In your code I cannot see you defining the variable outside the function. You are just echoing it out at the bottom of your code.

Answer 3

Any specific reason for keeping global variable inside method ?? Move it outside method and it should work for u.
Or
Try GLOBALS as shown below.
function doit() { $GLOBALS['val'] = 'bar'; } doit(); echo $val;
Gives the output as :
bar
Or
<?php
foo();
bar();
function foo() { global $jabberwocky; $jabberwocky="test data<br>"; bar(); } function bar() { global $jabberwocky; echo $jabberwocky; } ?>