Question

我不明白为什么我删除第14行中的代码（$ this-＆gt; close（）;），它没有错误但是我没有删除它然后警告mysqli_query（）：不能取mysqli。它是在结尾构造??? 我的错误：enter image description here 我的代码：

<?php

class Display extends Awebarts
{
    private $conn;
    private $tablename;

    public function __construct($tablename)
    {

        $this->tablename = $tablename;

        $this->connectToDb();
        $this->conn = $this->getConn();
        // insert the data into the table
        $this->getData();

        $this->close();
    }

    function getData()
    {
        $query = "SELECT * FROM $this->tablename ORDER BY `id` DESC LIMIT 1 ";
        if(!$sql = mysqli_query($this->conn, $query)) {
            throw new Exception("Error: Can not excute the query.");
        } else {
            $num = mysqli_num_rows($sql);
            while($num > 0) {
                //var_dump($data);
                $data = mysqli_fetch_array($sql);
                $num--;
            }
        }
        return $data;
    }
}

class Awebarts
{
    private $cxn;
    private $conn;

    function connectToDb()
    {
        include "models/Database.php";
        $vars = "include/vars.php";
        $this->cxn = new Database($vars);
        $this->conn = $this->cxn->getConn();
    }

    function getConn()
    {
        return $this->conn;
    }

    function close()
    {
        $this->cxn->close();
    }
}

class Database
{
    private $host;
    private $user;
    private $password;
    private $database;
    private $conn;

    function __construct($filename)
    {
        if(is_file($filename)) {
            include $filename;
        } else {
            throw new Exception("Error");
        }
        $this->host = $host;
        $this->user = $user;
        $this->password = $password;
        $this->database = $database;
        $this->connect();
        $this->selectData();
    }

    function getConn()
    {
        return $this->conn;
    }

    private function connect()
    {
        // connect to the sercer
        if(!mysqli_connect($this->host, $this->user, $this->password)) {
            throw new Exception("Error: not connected to the server");
        } else {
            $this->conn = mysqli_connect($this->host, $this->user, $this->password);
        }
        return $this->conn;
    }

    private function selectData()
    {
        if(!mysqli_select_db($this->conn, $this->database)) {
            throw new Exception("Error: No database");
        }
    }

    function close()
    {
        mysqli_close($this->conn);
    }
}

?>

Answer 1

尝试执行方法Database :: connect，如下所示：

private function connect()
        {
            // connect to the sercer
            if(!$connect = mysqli_connect($this->host, $this->user, $this->password)) {
                throw new Exception("Error: not connected to the server");
            }
            $this->conn = $connect;
            return $this->conn;
        }

Answer 2

问题是，在您创建getData()对象后调用Display时，连接已关闭。

构造函数中的getData()调用没有任何意义，因为您没有使用/保存返回值。因此，当执行构造函数时，您打开一个连接，发送一个选择查询（返回值，您不保存），然后关闭连接。之后，getData()调用会导致出现错误消息。

您可以将私有字段中构造函数的getData()调用结果保存并稍后访问，也可以从构造函数中删除getData()和$this->close();调用，然后从外部。

为什么警告：mysqli_query（）：无法获取mysqli？

2 个答案: