Question

这是我的代码：

if (isset($_POST['addmonths'])){
    if (!empty($_POST['months'])){
        if (is_numeric($_POST['months'])){
            $monthtoadd = $_POST['months'];
            if ($monthstoadd == 0){
                mysql_query("UPDATE users SET months='lifetime' WHERE username='$lookupuser'");
                echo "Successfully set " . $lookupuser . " to lifetime" . $monthstoadd;
            }elseif ($monthstoadd > 0){
                $monthstoadd = $monthstoadd*2592000;
                mysql_query("UPDATE users SET months=months+'$monthstoadd' WHERE username='$lookupuser'");
                echo "Successfully added " . $monthstoadd . " months to " . $lookupuser . "'s paid time.";
            }else{
                echo "Error.";
            }
        }else{
            echo "Months need to be numeric. If you're trying to set lifetime, use 0.";
        }
    }else{
        echo "You didn't enter anything.";
    }
}

无论我输入的是什么号码，它似乎总是将其设置为生命周期，然后它不会回显$monthstoadd后面的那个，这就是帮助我看看为什么它不起作用。我无法想象我的生活。如果我没有输入任何东西，它会像它应该的那样回复You didn't enter anything.，如果它不是数字，它就像它应该的那样回应Months need to be numeric. If you're trying to set lifetime, use 0.。

有人有什么想法吗？

Answer 1

第4行：

$monthtoadd = $_POST['months'];

应该是$ monthstoadd（注意缺失的s）

Answer 2

$monthtoadd = $_POST['months'];

应该是

$monthstoadd = $_POST['months'];

为什么我的变量似乎总是为0？

2 个答案: