开发者

PHP function wont save variables

开发者 https://www.devze.com 2023-03-04 23:09 出处:网络
I have a function that looks like this: function findSchedule($team) { switch($team) { case \"Baltimore Orioles\":

I have a function that looks like this:

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

        $team_home[42] = "Tampa Bay Rays";
        $team_home[43] = "Boston Red Sox";
        $team_home[44] = "Boston Red Sox";
        $team_home[45] = "$team";
        $team_home[46] = "$team";
        $team_home[47] = "$team";
        $team_home[48] = "$team";

        $team_away[42] = "$team";
        $team_away[43] = "$team";
        $team_away[44] = "$team";
        $team_away[45] = "New York Yankees";
        $team_away[46] = "New York Yankees";
        $team_away[47] = "Washington Nationals";
        $team_away[48] = "Washington Nationals";

        $team_date[42] = "Sun, May 15";
        $team_date[43] = "Mon, May 16";
        $team_date[44] = "Tue, May 17";
        $team_date[45] = "Wed, May 18";
        $team_date[46] = "Thu, May 19";
        $team_date[47] = "Fri, May 20";
        $team_date[48] = "Sat, May 21";

        break;

        case "Boston Red Sox":

        $team_home[42] = "$team";
        $team_home[43] = "$team";
        $team_home[44] = "$team";
        $team_home[45] = "$team";
        $team_home[46] = "$team";
        $team_home[47] = "$team";
        $team_home[48] = "$team";

        $team_away[42] = "Baltimore Orioles";
        $team_away[43] = "Baltimore Orioles";
        $team_away[44] = "Detroit Tigers";
        $team_away[45] = "Detroit Tigers";
        $team_away[46] = "Chicago Cubs";
        $team_away[47] = "Chicago Cubs";
        $team_away[48] = "Chicago Cubs";

      开发者_运维百科  $team_date[42] = "Mon, May 16";
        $team_date[43] = "Tue, May 17";
        $team_date[44] = "Wed, May 18";
        $team_date[45] = "Thu, May 19";
        $team_date[46] = "Fri, May 20";
        $team_date[47] = "Sat, May 21";
        $team_date[48] = "Sun, May 22";

        break;
  }

  for($i = 42;$i < 49;++$i)
  {
    return $team_home[$i];
    return $team_away[$i];
    return $team_date[$i];
  }

When I try to use the $team_date, $team_away, and $team_home variables as follows, the only one that seems to work is the $team_home variable.

$game = filter_input(INPUT_GET, 'game', FILTER_SANITIZE_STRING);

$team_home[$game] = findSchedule($team);
$team_away[$game] = findSchedule($team);
$team_date[$game] = findSchedule($team);

Any ideas?

Thanks,

Lance


It has to do with how you constructed your final for loop. You have it set to return each value; the return statement ends processing. You want to echo each one and then return, or alternatively never explicitly state return but just end the function.

To return all arguments, you could just make a single, larger array and return that:

return array( $team_home, $team_away, $team_date);


On the returning side you write an array:

return array($team_home, $team_away, $team_date);

And on the receiving side you can use list():

list($team_home, $team_away, $team_date) = findSchedule($team);
0

精彩评论

暂无评论...
验证码 换一张
取 消