一个PHP比较简单的有关问题

程序员文章站 2022-05-24 12:06:30

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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."
"; 

	?>