PHP Mysqli 准备语句 - 更新不更新

Question

在此先感谢大家的帮助。我有一个问题让我发疯了几个小时。我已经尝试使用几种具有完整错误检查的方法多次编写此代码。我试过在字段名称上加上单引号，我试过将 SQL 查询放在变量中并将其传递给 $db->prepare —— 都无济于事。我检查了我的权限，对我来说一切都很好。我敢肯定这很简单，但我已经睡眼惺忪地看了一遍，只是没有看到它。

$db = OpenDBConn();
    // $query  = "UPDATE agent_profiles SET ";
    // $query .= "website = ?, display_email = ?, primary_phone = ?, secondary_phone = ?, secondary_phone_type = ?, ";
    // $query .= "address_1 = ?, address_2 = ?, city = ?, state = ?, zip = ?, country = ?, description = ? ";
    // $query .= "WHERE agent_id = ?";
    $stmt = $db->prepare("UPDATE agent_profiles SET 
        website=?,
        display_email=?,
        primary_phone=?,
        secondary_phone=?,
        secondary_phone_type=?,
        address_1=?,
        address_2=?,
        city=?,
        state=?,
        zip=?,
        country=?,
        description=?
        WHERE agent_id=?");
    $stmt->bind_param('ssssssssssssi', $this->website, $this->display_email, $this->primary_phone, $this->secondary_phone, $this->secondary_phone_type, $this->address_1, $this->address_2, $this->city, $this->state, $this->zip, $this->country, $this->description, $this->agent_id);
    $stmt->execute();
    $stmt->close();
    $db->close();

即使有完整的错误报告并修改代码以查找 $db->e​​rror，它看起来和运行起来都很干净，但不会保存到表中。此处使用的功能在其他地方使用并且工作正常。有什么猜测吗？

Answer 1

error_reporting(-1);
ini_set('display_errors', 'On');

是你的 friend ，为什么你总是在旅途中排斥他？