通过PHP中的OOP连接到数据库和活动

Question

嗨，我在使用OO PHP连接到我的数据库时遇到了一些问题。我的脚本如下。我已经有一段时间了。这只是一个测试脚本，因为我对OOP相当新。请不要苛刻

class Database{

    public $mysqli,
         $host,
         $username,
         $password,
         $db;

    public function __construct($host, $username, $password, $db){

      $this->host = $host;
      $this->username = $username;
      $this->password = $password;
      $this->db = $db;

     $mysqli = new mysqli($host, $username, $password, $db);
        if (!$mysqli){
            echo "error in connecting to database";
        }
        else{
            echo "success in connecting to database";
        }
    }   

    public function query(){
        $result = $mysqli->query("SELECT * FROM inventory");
        if ($result) {
            printf("Select returned %d rows.\n", $result->num_rows);

            $result->close();
        }
        else{
            echo "there is an error in query";
            $result->close();
        }
        //echo "in query function";
    }
}

用法...

$DB = new Database('localhost', 'root', 'xxxx', 'yyyy');
$DB -> query();  

Answer 1

您的主要问题是您没有将值存储到您的班级'$mysqli属性中。您需要在构造函数和查询方法中使用$this->mysqli而不是$mysqli。

其次，这个类没有增加mysqli类没有的东西。你也可以简单地使用

$DB = new mysqli('localhost', 'root', 'xxxx', 'yyyy');

if ($DB->connect_error) {
    throw new Exception($DB->connect_error, $DB->connect_errno);
}

Answer 2

问题是，您正在访问没有$this的类属性。 $mysqli应该在构造函数中的$this->mysqli中。

您正在使用mysqli属性作为mysqli类的对象。

然后你可以在查询函数中使用$this->mysqli->query("SELECT * FROM inventory")的mysqli属性对象（在数据库类中）的查询函数。

更新的代码：

<?php
class Database{

    public $mysqli,
         $host,
         $username,
         $password,
         $db;

public function __construct($host, $username, $password, $db){

  $this->host = $host;
  $this->username = $username;
  $this->password = $password;
  $this->db = $db;

 $this->mysqli = new mysqli($host, $username, $password, $db);

    /* check connection */
    if ($this->mysqli->connect_errno) {
       printf("Connect failed: %s\n", $this->mysqli->connect_error);
       exit();
    } else {
       echo "success in connecting to database";
    }
}   

public function query(){
    $result = $this->mysqli->query("SELECT * FROM inventory");
    if ($result) {
        printf("Select returned %d rows.\n", $result->num_rows);

        $result->close();
    }
    else{
        echo "there is an error in query";
        $result->close();
    }
    //echo "in query function";
}
}

$DB = new Database('localhost', 'root', 'xxxx', 'yyyy');
$DB -> query();  

Answer 3

试试这个：

$connection = mysqli_connect(DB_SERVER, DB_USER, DB_PASS, DB_NAME);

// Test if connection succeeded
if(mysqli_connect_error()){
    die("Database connection failed: " .
        mysqli_connect_error() .
        " (" . mysqli_connect_errno() . ")"
    );
}