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Php 尝试向我的类中的mysql发送第二个查询时出错_Php_Mysqli - Fatal编程技术网

Php 尝试向我的类中的mysql发送第二个查询时出错

Php 尝试向我的类中的mysql发送第二个查询时出错,php,mysqli,Php,Mysqli,我有这段代码,但当我试图对db进行第二次查询时,它崩溃了,为什么?下面是代码的一部分,在这里它崩溃了 if ($this->doRegister === true) { $db = DB::connect(); $stmt = $db->prepare('SELECT `user_id` FROM `users` WHERE `user_name` = ? OR `user_email` = ? LIMIT 1'); $stmt-&

我有这段代码,但当我试图对db进行第二次查询时,它崩溃了,为什么?下面是代码的一部分,在这里它崩溃了

if ($this->doRegister === true) {
        $db = DB::connect();

        $stmt = $db->prepare('SELECT `user_id` FROM `users` WHERE `user_name` = ? OR `user_email` = ? LIMIT 1');

        $stmt->bind_param('ss', $this->store['userData']['name'], $this->store['userData']['email']);
        $stmt->execute();
        $stmt->bind_result($userId);
        $stmt->fetch();

        if (is_numeric($userId)) {
            $stmt = $db->prepare('INSERT INTO `users`(`user_name`, `user_password`, `user_email`, `user_ip`, `user_dateRegistered`, `user_type`) VALUES (?, ?, ?, ?, ?, ?)');

            $hashedPassword = $this->encrytion('md5', md5($this->store['userData']['name']) . md5($this->store['userData']['password']));
            $dateRegistered = time();
            $type = 1;

            $stmt->bind_param('ssssii', $this->store['userData']['name'], $hashedPassword, $this->store['userData']['email'], $_SERVER['REMOTE_ADDR'], $dateRegistered, $type);
            $stmt->execute();

            $this->registerUser();
        } else {
            return array('register' => 'User name or email already exists');
        }
    } else {
        return $this->store['userDataState'];
    }

在执行/准备新查询之前,需要关闭它

$stmt->close();

在执行/准备新查询之前,需要关闭它

$stmt->close();

在执行第二个stmt之前,您不需要关闭stmt吗?那太棒了!谢谢你,我爱你:在执行第二个stmt之前,你不需要关闭stmt吗?那太好了!谢谢你,我爱你