php - mysqli准备语句错误 "MySQL server has gone away"

Question

我正在努力实现从过程式到面向对象样式的跳转，所以如果我的代码不整洁或有缺陷，请多多关照 - 这里我通过 jQuery 将几篇文章传递给一个类，以便在用户更新记录时检查一个复选框:

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 prepare 语句并直接运行查询(注释掉)时，它会更新，当我尝试将其作为 prepare 语句运行时，它会返回错误“MySQL 服务器已消失”。

Answer 1

我查看了您的代码并找到了这一点。

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

我写的差不多。但是我看到你把连接和 prepare 函数分开了。

$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。