Question

此查询运行时没有错误，但它没有向数据库添加任何内容我无法弄明白我到处都看。你们有什么感想？我的数据库连接在另一个文件中定义，它可以工作，因为其他查询仍然有效。

  function processs_form(){

            $clean_fullname = sanitize($_POST['fullname']);
            $clean_sport = sanitize($_POST['sport']);
            $clean_package = sanitize($_POST['package']);
            $clean_weight = sanitize($_POST['weight']);
            $clean_height = sanitize($_POST['height']);
            $clean_gpa = sanitize($_POST['GPA']);
            $clean_number = sanitize($_POST['number']);
            $clean_school = sanitize($_POST['school']);
            $clean_gradYear = sanitize($_POST['gradYear']);
            $clean_yearRecorded = sanitize($_POST['yearRecorded']);
            $clean_postion = sanitize($_POST['position']);
            $clean_info = sanitize($_POST['additioninfo']);
            $reply_value = 1;
            GLOBAL $mysqli;

            $stmt = $mysqli->prepare("INSERT INTO video_tickets (`ticket_id`, `user_name`, `new_reply`, `full_name`, `sport`,
                    `position`, `year_recorded`, `grad_year`, `school`, `player_number`, `gpa`, `height`, `weight`,
                    `package`, `addition_info`) VALUES ( ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)");

            $stmt->bind_param('isisssiisidddss', $loggedInUser->user_id, $loggedInUser->username, $reply_value,
                    $clean_fullname, $clean_sport , $clean_postion, 
                    $clean_yearRecorded, $grad_year, $clean_school, $clean_number, $clean_gpa, 
                    $clean_height, $clean_weight, $clean_package, $clean_info);
                $stmt->execute();
                $stmt->close();


            $message = <<<_HTML_

            Your Ticket Has Been submited You'll get a Reply within 24 Hours.


_HTML_;

            print nl2br(htmlentities($message));




}

Answer 1

您的绑定已$grad_year但您使用$clean_gradYear作为变量：

$clean_gradYear = sanitize($_POST['gradYear']);

将bind_param值中的$grad_year更改为$clean_gradYear。

如果一个失败，您的整个查询都将失败。

错误报告会发出错误信号：

的 http://php.net/mysqli_error

Answer 2

这是一个很好的例子，说明为什么检查prepare，execute，bind_param等的返回值很重要。

您的查询很可能实际上失败了。它可能在bind_param阶段，正如其他人所指出的那样 - 你绑定的一个参数是未初始化的。

例如，execute()向呼叫者返回bool，表示成功或失败。您可以按如下方式检查：

if (!$stmt->execute()) {
    echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;

有关更完整的指南：PHP Prepared Statements。

查询子提交但不会添加到我的数据库

2 个答案: