Question

我有2个类，控制器和连接器。它们由delete.php使用。我收到错误说

Warning: mysqli_query() expects parameter 1 to be mysqli, null given in             C:\xampp\htdocs\pracownia\kontroler.php on line 38
Warning: mysqli_error() expects parameter 1 to be mysqli, null given in C:\xampp\htdocs\pracownia\kontroler.php on line 44

有没有人知道我犯了错误？对我来说，连接的链接看起来很糟糕地传递给某个地方的控制器类。或者回报不能按我的意愿运作。在我不得不把它变成课堂之前，这一切都做了......

类连接器在这里：

<?php

class connector {

private $db_handle="";
private $user_name;
private $password;
private $database;
public static $server = "127.0.0.1";

function __construct($user_name="", $password="", $database="") 
{
    $this->user_name = $user_name;
    $this->password = $password;
    $this->database = $database;
}

function connect() 
{
    $this->db_handle = mysqli_connect(self::$server,$this->user_name,$this->password);

if(!$this->db_handle)
{
    die('Could not connect: ' . mysqli_error());
}

    mysqli_select_db($this->db_handle,$this->database);
    return $this->db_handle;
}

function disconnect() 
{
    mysqli_close($this->db_handle);
}

}
?>

类控制器就在这里

<?php

class kontroler {

private $db_handle;
private $name;
private $surname;
private $password;
private $email;
private $id;
private $sql;

function _construct($db_handle="") 
{
    $this->db_handle = $db_handle;
}   

function delete_user($id="")
{
//$this->sql = "DELETE FROM UZYTKOWNICY WHERE ID_UZ=$id";

if(mysqli_query($this->db_handle,"DELETE FROM UZYTKOWNICY WHERE ID_UZ=$id")) 
    {
        echo "Record deleted successfully";
    }
         else 
    {
            die ('Error deleting record: ' . mysqli_error($this->db_handle));
    }
}


}

?>

和delete.php：

<?php
        global $db_handle;
        include 'kontroler.php';
        include 'connector.php';
        //require_once('kontroler.php');
        //require_once('connector.php');

        $connection = new connector("root","","proj");
        $db_handle = $connection->connect();
        $kontrol = new kontroler($db_handle);
        $kontrol->delete_user('$_POST[id]');
        $connection->disconnect();


    ?>

Answer 1

这是一个非常简短的答案，但我希望这会有所帮助。假设您的连接类位于＆＃39; connection.php＆＃39;，

class connector {

    private $db_handle; //db handles are not strings. Unsure why you were setting this to a string by default.
    private $user_name;
    private $password;
    private $database;
    public static $server = "127.0.0.1"; 
    //I'd recommend against this, unless you know 100% that your live server will use 'localhost'

    function __construct($user_name="", $password="", $database="") 
    {
        $this->user_name = $user_name;
        $this->password = $password;
        $this->database = $database;
        //There really isn't a need to not have your connector here - but that's me being lazy.
        $this->db_handle = mysqli_connect(self::$server,$this->user_name, $this->password, $this->database);

    }

    function query($sql){
        //Do some checks, or if you use prepared statements, accept the prepared data array and parse it here.
        return mysqli_query($this->db_handle, $sql);
    }

    function __destruct() 
    {
        mysqli_close($this->db_handle);
    }

}

然后，在您的页面中，执行以下操作：

include 'connector.php';

$db = new connector('user','pw','some_db');
$query = $db->query('DELETE FROM table WHERE user_id='.$_POST['something']);
//FOR THE LOVE OF GOD, DO NOT ACTUALLY DO THIS. I DID IT ONLY FOR THE SAKE OF BREVITY. USING RAW DATA SUBMITTED FROM THE USER
// IS DANGEROUS AND LEADS TO SQL INJECTIONS. *ALWAYS* SANITIZE YOUR DATA.

原因是在mysqli_connect和mysqli_close方法中添加__construct和__destruct调用是因为我非常懒惰而且不想打电话给->connect()方法。说实话，我的懒惰有时为我节省了很多工作，因为我不想一遍又一遍地重复自己（发现这是一种叫做DRY的编码哲学 - 不要重复自己）。希望这会有所帮助。

PHP mysqli_query（）期望参数1为mysqli，null为null

1 个答案: