Should be simple: function return error, what am I missing?

[Q695]

inside function:

echo "<br>random path: $path_rand";
        return $path_rand;

 

outside function:

echo "<p>random path2: $path_rand";

 

$path_rand is echo in the function, but for some reason not being posted outside of the function.  Why?

 

I tried the following simple script:

function test(){
    $i=2;
    return $i;
    }
    test();
    echo "X:$i";
 

 

got back: Notice: Undefined variable: in ______ on line X(echo line)

X:

[Barand]

read up on variable scope

 

http://php.net/manual/en/language.variables.scope.php

[Q695]

I checked the function return when it was erroring the first time, but i'll try again.

[Barand]

But you aren't using the value returned by the function - you are using a variable that was defined inside the function. As I said - variable scope

[Q695]

Should be a global, not a return

[Barand]

Should be a global, not a return

 

 

 

That reply deserves top spot in this site's Hall of Shame.

 

You should NOT use globals. You should pass values to functions then use the value returned by the function.

[Q695]

Where can I find the hall of shame for responses?