OOP mysqli数据库功能

Question

我是PHP OOP的初学者，并且尝试使用我的代码。 因此，我想做的第一件事是编写一个具有数据库连接和插入功能的基本类。 所以我想要的情况是这样的：

-我想在课堂上创建它将控制连接和插入函数的类。 问题是，我的$ connect变量无法通过另一个函数工作，那么我该怎么做才能做到这一点？ 您将通过提供的代码了解更多信息。

<?php

class DB {
protected $dbhost = 'localhost';
protected $dbuser = 'root';
protected $dbpass = '';
protected $dbname = 'newdb';

public function connect()
{
    $connect = new mysqli($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname);

    if($connect->error)
    {
        echo "Failed to connect";
    }
    else 
    {
        echo "connected";
    }
}

public function insert($name, $second)
{
    $insert = "INSERT INTO posts (name, second) VALUES ('$name', '$second')"; 

    if ($connect->query($insert) === TRUE) 
    {
        echo "New record created successfully";
    } 
    else 
    {
        echo "Error: " . $sql . "<br>" . mysqli_connect_error();
    }
}
}
require_once 'classes/DB.php';

$db = new DB;

if(isset($_POST['submit']))
{
    $name = $_POST['name'];
    $second = $_POST['second'];

    $db->insert($name, $second);
}

?>


<form method="POST">
    Add smth<input type="text" name="name"><br>
    Also omg<input type="text" name="second"><br>
    <input type="submit" name="submit">
</form>

Answer 1

将$connect添加$connect类的属性，因此您可以使用$this->connect在类中的任何地方重用它：

class DB
{
    protected $dbhost = 'localhost';
    protected $dbuser = 'root';
    protected $dbpass = '';
    protected $dbname = 'newdb';
    protected $connect;

    public function connect()
    {
        $this->connect = new mysqli($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname);

        if ($this->connect->error)
        {
            echo "Failed to connect";
        }
        else
        {
            echo "connected";
        }
    }

    public function insert($name, $second)
    {
        $insert = "INSERT INTO posts (name, second) VALUES ('$name', '$second')";

        if ($this->connect->query($insert) === TRUE)
        {
            echo "New record created successfully";
        }
        else
        {
            echo "Error: " . $sql . "<br>" . mysqli_connect_error();
        }
    }
}

Answer 2

我已经重写了您的类，因此您可以在构造函数中插入数据库设置。 因此，它确实是一个对象，您可以使用多个数据库。 您在类中做错的是使用变量$ connect而不将其声明为可在整个类中使用的变量。

班级

<?php

class DB {
    protected $dbhost;
    protected $dbuser;
    protected $dbpass;
    protected $dbname;
    protected $connection;

    public function __construct($dbhost, $dbuser, $dbpass, $dbname)
    {
        $this->dbhost = $dbhost;
        $this->dbuser = $dbuser;
        $this->dbpass = $dbpass;
        $this->dbname = $dbname;

        $connection = new mysqli($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname);
        if($connection->error)
            die('Could not connect with the database!');

        $this->connection = $connection;
    }

    public function connect()
    {
        $this->__construct($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname);
    }

    public function insert($name, $second)
    {
        $insert = "INSERT INTO posts (name, second) VALUES ('$name', '$second')"; 

        if ($this->connection->query($insert) === TRUE) 
        {
            echo "New record created successfully";
        } 
        else 
        {
            echo "Error: " . $sql . "<br>" . mysqli_connect_error();
        }
    }

    public function getConnection()
    {
        return $this->connection;
    }
}

用法

$db = new DB('localhost', 'root', '', 'newdb');
$db->insert('name', 'second');

我希望这可以通过OOP对您的日记有所帮助，对不起，我的英语不好。