PHP如何使用多个参数并用它们更新mySQL表

Question

我想用 PHP 中的以下函数更新表。问题是第二个参数$work_place不被接受，更新失败。这是我第一次使用 PHP 和 mySQL，所以我的知识有点有限。

public function timestampOut($work_done, $work_place)
    {
        // clean the input to prevent for example javascript within the notes.
        $work_done = strip_tags($work_done);
        $work_place = strip_tags($work_place);


        $userLastTimestampID = $this->getUserLastTimestampID();

        $sql = "UPDATE timestamps SET timestamp_work_description = :work_done, timestamp_work_dropdown = :work_place, timestamp_out = now() WHERE timestamp_id = $userLastTimestampID[0] AND user_id = :user_id";
        $query = $this->db->prepare($sql);
        $query->execute(array(':work_done' => $work_done, ':user_id' => $_SESSION['user_id']));



        $count =  $query->rowCount();
        if ($count == 1) {
            return true;
        } else {
            $_SESSION["feedback_negative"][] = FEEDBACK_NOTE_CREATION_FAILED;
        }
        // default return
        return false;
    }

Answer 1

您只需将 work_place 添加到执行调用中的 param 数组中，如下所示:

$query->execute(array(':work_done' => $work_done, ':work_place' => $work_place, ':user_id' => $_SESSION['user_id']));