一个PHP比较简单的有关问题

程序员文章站 2024-04-05 08:25:54

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求助一个PHP比较简单的问题
我想获取数据库受影响的行数可是$row怎么在实现类这边获取返回影响的行数呢，新手伤不起，这个弯不会转了一个PHP比较简单的有关问题



	// class_user.php 
class UserInfo{ 
   private $userName;  //属性,用户名 
   private $passWord;  //属性, 
   private $userGrade; //属性,用户级别
   private $row; 
   private $userInfo; //存储数据库返回信息的数组变量. 

	    public function __construct($sql){ 
	        require_once("db_config.php"); //包含配置信息. 
	        $result= mysql_query($sql, $conn); // 执行查询语句 
	        $this->userInfo = mysql_fetch_array($result); //返回查询结果到数组
			$row=mysql_num_rows($result); 
			mysql_query("set names UTF8");
	        mysql_close($conn); //关闭数据库连接 
	      $this->getInfo(); //调用传递信息的方法.	 
	    } 
	    // 获取信息传递给属性的方法 
	    private function getInfo(){ 

	        $this->userName = $this->userInfo["username"]; 
	        $this->passWord = $this->userInfo["password"];	
	        $this->userGrade = $this->userInfo["userGrade"]; 
	    } 

	    //返回每个属性的public 方法. 

   public function getRow(){ 

	        return $this->row; 
	    } 
	    public function getUserName(){ 

	        return $this->userName; 
	    } 

	    public function getPassword(){ 
	        return $this->passWord; 
	    } 


	    public function getUserGrade(){ 
	        return$this->userGrade; 
	    } 


	} 
		$sql= "select * from userinfo where username='sa' ";
	require_once("class_user.php");   
	$user= new UserInfo($sql); //创建一个user对象. 
	    //书写sql
	$row= $user->getRow(); 
	$username= $user->getUserName(); //分别调用方法取得数据 
	$password= $user->getPassword(); 
	$usergrade= $user->getUserGrade(); 

	echo"your name is ".$username."
";  //输出数据 
	echo"your password is ".$password."
"; 

	echo"your grade is ".$usergrade."
"; 
	echo"your ROW is ".$row."
"; 
	?>