PHP 和 MySQLi - 无法通过引用传递参数 2

PHP and MySQLi - Cannot pass parameter 2 by reference in(PHP 和 MySQLi - 无法通过引用传递参数 2)

问题描述

我正在尝试创建一个函数来检查更新并插入一些数据，但我在第一步中遇到了一个问题，其中 $stmt->bind_param 表示不是通过引用或类似方式传递参数.

I am trying to make a function which will check update and insert some data but I am having an issue in the first step where the $stmt->bind_param is saying that is not passing parameters by reference or something like that.

我在下面附上了功能代码:

I have attached below the function code:

public function killTarget($killerid,$victimiid,$victimcode)
    {

        if ($this->checkUsercode($victimcode,$victimiid))
        {
            $stmt = $this->_db->prepare("UPDATE users SET status =? WHERE user_id =?");
            $stmt->bind_param("ii",0,$victimiid);

            if ($stmt->execute())
            {
                $stmt->store_result();
                $stmt->fetch();

                $stmt = $this->_db->prepare("SELECT victim_id FROM target WHERE killer_id = ?");
                $stmt->bind_param("i",$victimiid);

                if ($stmt->execute())
                {
                    $stmt->store_result();
                    $stmt->bind_result($targetid);
                    $stmt->fetch();

                    $stmt = $this->_db->prepare("INSERT INTO target (killer_id, victim_id) VALUES (?,?)");
                    $stmt->bind_param("ii",$killerid,$targetid);

                    if ($stmt->execute())
                    {
                        $stmt->store_result();
                        $stmt->fetch();
                        $stmt->close();
                    }
                }
            }
            else
            {
                Main::setMessage("targets.php",$this->_db->error,"alert-error");
            }
        }

    }

推荐答案

你不能在mysqli中这样做:

You cannot do this in mysqli:

$stmt->bind_param("ii",0,$victimiid);

0 需要是一个变量.

试试这个:

$zero = 0;
$stmt->bind_param("ii",$zero,$victimiid);

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