Call Array Value From Outside Function?

[Mink_]

I'm working on a web-RPG, and need a little help with my database queries, as I need to be setting up globals using unfctions.

I'm using this simple code that I came up with after a few minutes of looking at the PHP manual (I've already set the database up):

PHP Code:

function getuserinfo($id) {
    $query = mysql_query("SELECT * FROM users WHERE id = $id LIMIT 1");
    $user = mysql_fetch_array($query);
} 


Then, I want to call values from that by using things like $user['var'].
The only problem is, I can only seem to call values from that array inside the function, meaning that this will work:

PHP Code:

function getuserinfo($id) {
    $query = mysql_query("SELECT * FROM users WHERE id = $id LIMIT 1");
    $user = mysql_fetch_array($query);
    echo $user['var'];
}
getuserinfo(1); 


And this will not:

PHP Code:

function getuserinfo($id) {
    $query = mysql_query("SELECT * FROM users WHERE id = $id LIMIT 1");
    $user = mysql_fetch_array($query);
}
getuserinfo(1);
echo $user['var']; 


Is there an alternate way of getting the results I would see if the second bit of code worked? Or do I have to redesign my entire system?

[amykhar]

You could set up the user variable outside the function and pass it in to be filled or you could return the user variable. The way you have it set up now, its scope is only within the function.

[Mink_]

You could set up the user variable outside the function and pass it in to be filled or you could return the user variable.

The reason I wanted to do this with arrays is that I need to have something like 100 global variables for my software, and setting them up individually would be a pain. They're taken directly from the MySQL database, and then turned into arrays, meaning I couldn't really set up that value elsewhere...

[amykhar]

You can pass arrays in and out just like any other variable.

[Mink_]

You can pass arrays in and out just like any other variable.

I'm having trouble passing the variables out. oO;
for instance,

PHP Code:

function test() {
$var = "val";
}
test();
echo $var; 


doesn't work.

[amykhar]

function test should have a return line in it that returns $var.

The code would look like:

function test() { 
$var = "val"; 
return $var;
} 
$othervar = test(); 
echo $othervar;

[Mink_]

Thanks alot.
You rule. This will help me infinitely. As the English say, cheers!