在 [重复] 中在 null 上调用成员函数 prepare()

Question

这是我的代码，我遇到了这个错误

在 null in 上调用成员函数 prepare() 我认为我的课程没问题，但有一个错误，我不知道如何让这段代码运行

<?php
/*$servername = "localhost:3307";
$username = "root";
$password = "";
$dbname = "motorway";*/

class BaseConnect {
    private $conn;

    function __construct() {
    }

    // Create connection
    function connect() {
        $conn = new mysqli("localhost:3307", "root", "", "motorway");
        // Check connection
        if ($conn->connect_error) {
            die("Connection failed: " . $conn->connect_error);
        }
        return $this->conn;
    }
}
class Database {
    private $conn;

    function __construct() {
        $db = new BaseConnect();
        $this->conn = $db->connect();
    }

    function Enterdata($name, $phone, $email, $password) {
        $pass = md5($password);
        $sql = $this->conn->prepare("INSERT INTO 'user'('name', 'phone', 'email', 'pass')
    VALUES(?, ?, ?);");
        $sql->bind_param("ssss", $name, $phone, $email, $pass);
        if ($sql->execute()) {
            return 1;
        } else {
            return 2;
        }
    }
}

Answer 1

如果您想使用私人$conn 成员，您必须更改此代码：

$conn = new mysqli("localhost:3307", "root", "", "motorway");
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

用这个：

$this->conn = new mysqli("localhost:3307", "root", "", "motorway");
// Check connection
if ($this->conn->connect_error) {
    die("Connection failed: " . $this->conn->connect_error);
}

无论如何BaseConnect 类是完全多余的，你应该删除它并使用Database 类。