这是什么错误？

Question

class MysqlConn
{
	
	public $dbtype;
	public $host;
	public $dbname;
	public $user;
    public $pwd;

	function __construct($dbtype,$dbname,$host,$user,$pwd)
	{
			$this->dbtype=$dbtype;
			$this->host=$host;
			$this->dbname=$dbname;
			$this->user=$user;
			$this->pwd=$pwd;
			if($this->dbtype=="mysql" || $this->dbtype=="mssql")
			{
    		    $dsn="$this->dbtype:host=$this->host;dbname=$this->dbname";
		    }
			else
			{
			    $dsn="$this->dbtype:dbname=$this->dbname";
		    }
			try 
			{
				$conn = new PDO($dsn, $this->user, $this->pwd); 	//初始化一个PDO对象，就是创建了数据库连接对象$pdo
				$conn->query("set names utf8");
				return $conn;
		    } 
			catch(PDOException $e)
			{
    		    die("Error!: " . $e->getMessage() . "<br/>");
		    }    
			
	} 
}


$conn=new MysqlConn("mysql","dzxx_db","localhost","admin","123");//数据库连接类实例化,执行连接操作，返回连接标识

Answer 1

$dsn的值有问题，正确的应该分别是 $dsn=$this->dbtype.":host=".$this->host.";dbname=".$this->dbname;和$dsn=$this->dbtype.":dbname=".$this->dbname;

Answer 2

class ConnDB{ var $dbtype; var $host;     var $user;     var $pwd;     var $dbname;      //构造方法     function ConnDB($dbtype,$host,$user,$pwd,$dbname){ $this->dbtype=$dbtype;           $this->host=$host; $this->user=$user; $this->pwd=$pwd; $this->dbname=$dbname; }     //实现数据库的连接并返回连接对象     function GetConnId(){            if($this->dbtype=="mysql" || $this->dbtype=="mssql"){      $dsn="$this->dbtype:host=$this->host;dbname=$this->dbname"; }else{ $dsn="$this->dbtype:dbname=$this->dbname"; }     try {      $conn = new PDO($dsn, $this->user, $this->pwd);  //初始化一个PDO对象，就是创建了数据库连接对象$pdo $conn->query("set names utf8");      return $conn; } catch (PDOException $e) {      die ("Error!: " . $e->getMessage() . "<br/>"); }           } }上面这个是原本的类，可没见需要像你那样写啊