好的,这是我的代码:
function payCalc($hours, $wage, $overtime, $paycheck) {
if ($hours <= 40) {
$paycheck = $hours * $wage;
}
else {
$paycheck = $hours * $overtime;
}
return $paycheck;
}
$name = $_POST['name'];
$dept = $_POST['dept'];
$hours = $_POST['hours'];
$wage = $_POST['wage'];
$paycheck = payCalc($hours, $wage, $overtime, $paycheck);
$overTime = $wage * 1.5;
print "Your paycheck for this period is:";
print $paycheck;
print ".";
问题是我在 $paycheck 上不断收到未定义的变量错误。据我所知,我已经正确设置了 if/else,但它没有返回$paycheck
。这意味着在“ print $paycheck;
”行上,它无法打印。我错过了什么?
编辑:还有更多内容,但这只是吐出$name
and$dept
变量的回声线。