PHP OOP -如何将mysqli方法继承到其他类


PHP OOP - How to inherit mysqli methods to other class?

所以我想做一个简单的方法来管理一堆mysql连接。

我有两个基本类:serverConnections

服务器只是设置为保存一行数据,如您所见。(有更好的方法吗?)

在连接中,我想通过它的ID调用连接并创建连接,并能够查询它。

这与mysql库一起工作,但是当我尝试使用mysqli库时,它不会将$mysqli传递给$msi作为连接句柄。$mysqli->query()可以工作,但$msi->query()不行。

   PHP Fatal error:  Call to undefined method Connections::query()

我需要扩展/实现我的连接类与mysqli或以不同的方式返回$mysqli ?

谢谢!

    <?php
    class server {
        public $id, $type, $subid, $host, $user, $pw, $port, $db;
        public function __construct($id,$type,$subid,$host,$user,$pw,$port,$db) {
            $this->id = $id;
            $this->type = $type;
            $this->subid = $subid;
            $this->host = $host;
            $this->user = $user;
            $this->pw = $pw;
            $this->port = $port;
            $this->db = $db;
        }
    }
    class Connections {
        public $servers;
        function __construct($id) {
            $this->servers[] = new server("mysql","mysql","main","hostname","username","password","3306","dbname");
            $this->servers[] = new server("mysql2","mysqli","main","hostname","username","password","3306","dbname");
            $rt = null;
            foreach($this->servers as $server) {
                if ($id == $server->id) {
                    $rt = $server;
                    break;
                }
            }
            if($rt->type == "mysql"){
                $con = mysql_connect($rt->host,$rt->user,$rt->pw);
                mysql_select_db($rt->db, $con);
                    if($con) { return $con; }
            }
            elseif($rt->type == "mysqli"){
                $mysqli = new mysqli($rt->host, $rt->user, $rt->pw, $rt->db, $rt->port);
                        if ($mysqli->connect_errno) {
                              echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
                        }
                        else{
                              return $mysqli;
                        }
            }
        }
    }
    /*
    ### this works
    $nc = new Connections("mysql");
    $q = "select 1+2 as asdf";
    $r = mysql_query($q);
    while($row = mysql_fetch_array($r)){
        echo $row['asdf']."'n";
    }
    */
    #### this does not work
    $msi = new Connections("mysql2");
    $res = $msi->query("SELECT 2+2 as asdf");
    while($row = $res->fetch_assoc()){
        echo $row['asdf']."'n";
    }
    ?>
#

EDIT ->在这里的另一个帖子的帮助下,我能够以不同的方式完成这项工作

这是我修改后的代码:

<?php
class server {
    public $id, $type, $subid, $host, $user, $pw, $port, $alt; 
    public function __construct($id,$type,$subid,$host,$user,$pw,$port,$alt) {
        $this->id = $id;
        $this->type = $type;
        $this->subid = $subid;
        $this->host = $host;
        $this->user = $user;
        $this->pw = $pw;
        $this->port = $port;
        $this->alt = $alt;
    }  
}
class MySQLiContainer extends SplObjectStorage{
    var $server, $servers;
    function __construct($id) {
        $this->servers[] = new server("mysql","mysql","main","hostname","username","password","3306","dbname");
        $this->servers[] = new server("mysql2","mysqli","main","hostname","username","password","3306","dbname");
         foreach($this->servers as $server) {
                if ($id == $server->id) {
                    $this->server = $server;
                    break;
                }
            }
    }
  public function newConnection() {
    $mysqli = new mysqli($this->server->host, $this->server->user, $this->server->pw, $this->server->alt, $this->server->port);
    $this->attach($mysqli);
    return $mysqli;
  }
}
//usage
$mysqliContainer = new MySQLiContainer("mysql2");
$c1 = $mysqliContainer->newConnection();
$res = $c1->query('SELECT 2+4 as asdf');
while($row = $res->fetch_assoc()){
    echo $row['asdf']."'n";
}
echo $c1->host_info . "'n";
$ms2 = new MySQLiContainer("mysql");
$c2 = $ms2->newConnection();
$r = $c2->query('SELECT 2+4 as dfdf');
while($ra = $r->fetch_assoc()){
    echo $ra['dfdf']."'n";
}
echo $c2->host_info . "'n";
?>

你的代码是基于一个错误的前提:你可以从构造函数返回任何东西。

构造函数不能返回任何东西。new操作符的结果始终是类的实例。

让我重新表述一下,以防不清楚。不能从构造函数返回任何内容,因为return的值将被丢弃。

你似乎想做的是 extendmysqli。

但是,您实际上不想这样做,主要有两个原因:

  1. mysqli不允许返回相关类,如语句句柄或结果句柄,作为派生类。这可能会严重限制功能。如果你真的想扩展一个数据库处理程序,使用PDO。你为什么要写自己的数据库抽象库?有几十个这样的。去看看Zend Db或者Doctrine用的什么。它们将对你更有用——由合格的第三方创建并经过良好测试。