我有以下 PHP。
最终,我希望我在执行结束时的输出是:
Size1Size3
$footballTypes = "1,3";
$footballTypeNames = "";
foreach ( $footballTypes as $data ) {
switch ($data) {
case 1:
$footballTypeNames .= "Size 1";
case 2:
$footballTypeNames .= "Size 2";
case 3:
$footballTypeNames .= "Size 3";
}
}
echo $footballTypeNames;
但是,现在,我收到错误消息:
Warning: Invalid argument supplied for foreach() in C:\xampp\htdocs\football.php on line 7
我哪里错了?
30秒到达战场
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