Question

我想用PHP中的following函数更新表。问题是不接受第二个参数$ work_place，更新失败。这是我第一次使用PHP和mySQL，所以我的知识有点受限。

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

您只需要在执行调用中将work_place添加到param数组中，如下所示：

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

Answer 2

请再次阅读execute的工作原理。您可能希望像这样使用它：

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

Answer 3

尝试替换

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

与

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

PHP如何使用多个参数并使用它们更新mySQL表

3 个答案: