將OOP PHP數據庫結果轉換為數組

Question

我是oop php的新手，我正在嘗試編寫一個irc php。

我想做的是：我想查詢數據庫，從數據庫中獲取結果，並將其放入程序內部的數組中。

我嘗試制作一個新函數來執行任務，並在__construct函數中對其進行了調用。

我已經縮短了代碼，但是看起來幾乎是這樣的：

任何想法都值得贊賞。

class IRCBot
{    

    public $array = array();
    public $servername = "localhost";
    public $username = "root";
    public $password = "usbw";
    public $dbname = "bot";


function __construct()
{
    //create new instance of mysql connection
    $conn = new mysqli($this->servername, $this->username, $this->password, $this->dbname); 

    if ($mysqli->connect_errno) 
    {
        echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
    }
    echo $mysqli->host_info . "\n";

    $this->database_fetch();

}

function database_fetch()
{
    $query = "SELECT word FROM timeoutwords";
    $result = mysqli_query($query);
    while($row = mysqli_fetch_assoc($result))
        {
            $array[] = $row();
        }
}

function main()
{
    print_r($array);
}



}

$bot = new IRCBot();

Answer 1

首先，您需要修復構造函數中的錯誤，您可以將其修改為：

function __construct()
{
    //create new instance of mysql connection
    $this->conn = new mysqli($this->servername, $this->username, $this->password, $this->dbname); 

    if ($this->conn->connect_errno) 
    {
        echo "Failed to connect to MySQL: (" . $this->conn->connect_errno . ") " . $this->conn->connect_error;
    }
    echo $this->conn->host_info . "\n";
}

• 在這里，您需要用$conn替換$mysqli ，因為您的鏈接標識符是$conn而不是$mysqli
• 無需在此處調用database_fetch() 。
• 您需要將$conn用作屬性。

現在，您需要將database_fetch()方法修改為：

function database_fetch()
{
    $query = "SELECT word FROM timeoutwords";
    $result = mysqli_query($this->conn,$query);
    $array = array();
    while($row = mysqli_fetch_assoc($result))
    {
        $array[] = $row;
    }
    return $array;
}

• 在這里，您需要在mysqli_query()傳遞添加第一個參數，該參數應該是鏈接標識符/數據庫連接。

• 其次，您需要使用return從此函數獲取結果。

最后，您需要將main()方法修改為：

function main()
{
    $data = $this->database_fetch();
    print_r($data);
}

• 在這里，您需要在此處調用database_fetch()方法，然后在所需的位置打印數據。

Answer 2

變化

1）將 if ($mysqli->connect_errno) ）更改為if ($conn->connect_errno)

2）更改$array[] = $row(); 到$array[] = $row;

3）添加return $array; 在函數database_fetch()

4）在main()函數而不是構造函數中調用database_fetch()函數。

5）在mysqli_query() ）中添加$this->conn （感謝@devpro指出）。

更新的代碼

<?php
class IRCBot
{    

    public $array = array();
    public $servername = "localhost";
    public $username = "root";
    public $password = "usbw";
    public $dbname = "bot";
    public $conn;


    function __construct()
    {
        //create new instance of mysql connection
        $this->conn = new mysqli($this->servername, $this->username, $this->password, $this->dbname); 

        if ($this->conn->connect_errno) 
        {
            echo "Failed to connect to MySQL: (" . $this->conn->connect_errno . ") " . $this->conn->connect_error;
        }

    }

    function database_fetch()
    {
        $query = "SELECT word FROM timeoutwords";
        $result = mysqli_query($this->conn,$query);
        while($row = mysqli_fetch_assoc($result)){
            $array[] = $row;
        }
        return $array;
    }

    function main()
    {
        $data = $this->database_fetch();
        print_r($data);
    }

}

快速開始

• PHP中的面向對象編程
• 類和對象
• PHP中的面向對象編程原理