Question

I have this php function that I created that is basically a switch statement. For each case, the $team_image variable is saved to a different value. It looks a little something like this:

function teamImage($team)
{
    switch($team)
    {
        case "Baltimore Orioles":

            $team_image = "orioles";

        case "New York Yankees":

            $team_image = "yankees";

        case "Toronto Blue Jays":

            $team_image = "bluejays";

You get the idea. However, when I call on the function and try to use the $team_image variable in other parts of my code, it doesn’t work because apparently, the variable is still undefined. any thoughts?

Thanks,

Lance

Answer 1

As you’re only setting the $team_image inside the teamImage function, it will only exist with that function’s “scope“. (In general, variables, etc. will always exist in as narrow a scope as possible, which is good in terms of encapsulation. (Encapsulation being a key benefit of object orientated programming, etc. which you may go on to discover as you learn more.)

As such, you should return the $team_image value from the teamImage function and set it as follows:

function teamImage($team) {

    $team_image = NULL;

    switch($team) {
       ...
    }    

    return $team_image;
}

$team_image = teamImage($team);

An alternative would be to define the $team_image variable within the teamImage function as a global by adding the line global $team_image; at the beginning of the function, but this isn’t considered good practice.

Additionally, you should break; each case code block within your switch statement as otherwise you’ll simply end up setting $team_image with the value assigned in your final case. (i.e.: If you don’t break each statement, code flow will continue into the next.) See the switch PHP manual page for full details.