Question

这是我连接数据库的类，但是当我创建该类的实例时，它给了我这个错误：第21行的C：\ Apache24 \ htdocs \ core \ DB.class.php中的Fatal error: Call to undefined method DB::query()

 <?php

class DB {
    protected $db_name = "data_db";
    protected $db_user = "root";
    protected $db_pass = "root";
    protected $db_host = "localhost";

    public function __construct() {
        $db = new mysqli($this->db_host, $this->db_user, $this->db_pass, $this->db_name);
        if ($db->connect_errno) {
            printf("Connection failed: %s\n", $db->connect_error);
            exit();
        } else
            return $db;
    }
}

$db = new DB();

$data_row = $db->query("SELECT * FROM `users`");
while($data = $data_row->fetch_assoc())
    echo $data["first_name"] . '<br>';

?>

致命错误：在第21行的C：\ Apache24 \ htdocs \ core \ DB.class.php中调用未定义的方法DB :: query（）

Answer 1

构造函数不返回自定义值，它们返回类的实例。要测试此使用var_dump( $db );，输出将为object(DB) ...。将连接保存为对象的属性，如：

public function __construct() {
    $this->mysqli = new mysqli($this->db_host, $this->db_user, $this->db_pass, $this->db_name);
}

$db = new DB();
$data_row = $db->mysqli->query("SELECT * FROM `users`");

从MySQL数据库获取结果时出错

1 个答案: