mysqli 准备语句错误“MySQL 服务器已消失" [英] mysqli prepare statement error "MySQL server has gone away"

问题描述

我正在努力制作程序到面向对象样式的跳转形式，所以如果我的代码不整洁或有缺陷，请保持友好 - 在这里，我通过 jQuery 将一些帖子传递给一个类，以便在用户使用时更新记录选中一个复选框:

I'm struggling to make the jump form Procedural to Object Orientated style so if my code is untidy or flawed please be nice - here I'm passing a couple of posts via jQuery to a class to update a record when the user checks a checkbox:

这里是数据库连接

class db {

    private $host ;      
    private $username;
    private $password;
    private $dbname;

    protected function conn()
    {
        $this->host = "localhost";
        $this->username = "root";
        $this->password = "";
        $this->dbname = "mytest";

        $db = new mysqli($this->host, $this->username,  $this->password, $this->dbname);

        if($db->connect_errno > 0){
                die('Unable to connect to database [' . $db->connect_error . ']');
        }

        return $db;

    }

}

这里是更新类

class updOrders extends db {

public $pid;
public $proc;

public function __construct()
{
$this->pid = isset($_POST['pid']) ? $_POST['pid'] : 0;
$this->proc = isset($_POST['proc']) ? $_POST['proc'] : 1;

// $stmt = $this->conn()->query("UPDATE tblorderhdr SET completed = ".$this->proc." WHERE orderid = ".$this->pid);

$stmt = $this->conn()->prepare("UPDATE tblorderhdr SET completed = ? WHERE orderid = ?");
$stmt->bind_param('ii', $this->proc, $this->pid);

 $stmt->execute();

if($stmt->error)
    {
        $err = $stmt->error ;
    } else {
        $err = 'ok';
    }

/* close statement */
$stmt->close();

 echo json_encode($err);   

}

}

$test = new  updOrders;

当我注释掉准备语句并直接运行查询(注释掉)时，它会更新，当我尝试将它作为准备语句运行时，它返回错误MySQL 服务器已消失".

When I comment out the prepare statement and run the query directly (commented out) it updates, when I try and run it as a prepare statement it returns an error "MySQL server has gone away".

非常感谢任何帮助我已经没有想法了！

Any help very much appreciated I've run out of ideas!

PS:本地主机上的 Apache/XAMPP

PS: Apache / XAMPP on localhost

推荐答案

我查看了你的代码，发现了这个.

I have looked at your code and I found this.

$stmt = $this->conn()->prepare("UPDATE tblorderhdr SET completed = ? WHERE orderid = ?");

我写的几乎一样.但是我看到你把连接和准备功能分开了.

I wrote nearly the same. But I saw you separated the connection from the prepare function.

$db = $this->conn();

$stmt = $db->prepare("UPDATE tblorderhdr SET completed = ? WHERE orderid = ?");

我也不知道为什么，但现在可以了.

I don't know either why, but it works now.

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