php - 在函数之间传递数据库连接

Question

我是 PHP 的新手，所以也许这是一个简单的问题。这是我用来更新数据库的类。问题是它一直在标记为* 的行给我一个错误，因为它找不到$con，这显然在函数openconn() 中。看来我无法将连接传递给另一个函数。我做错了什么吗？谢谢

class retreats {

    public $retreat_name = '';

    function openconn() {
        $con = mysql_connect("localhost","root","root");

        if (!$con)
        {
            die('Could not connect: ' . mysql_error());
        }
        mysql_select_db("PHPTest", $con);
    }


    function closeconn(){
        mysql_close($con);
    }


    function add_retreat(){
        openconn();
        $sql="INSERT INTO tbl_retreats (retreat_name) VALUES ('".$this->retreat_name."')";

        if (!mysql_query($sql,$con)) *******
        {
            die('Error: ' . mysql_error());
        }

        echo "Record Successfully Added";

        closeconn();

    }
}

Answer 1

$con 是函数 openconn 的局部变量。尝试以这种方式更改您的代码:

class retreats {  

    public $retreat_name = ''; 

    private $con;

    function openconn() {
    $this->con = mysql_connect("localhost","root","root");

    if (!$this->con)
    {
        die('Could not connect: ' . mysql_error());
    }
        mysql_select_db("PHPTest", $this->con);
    }


    function closeconn(){
        mysql_close($this->con);      
    }


    function add_retreat(){  
        openconn();
        $sql="INSERT INTO tbl_retreats (retreat_name) VALUES ('".$this->retreat_name."')"; 

        if (!mysql_query($sql,$this->con))
        {
            die('Error: ' . mysql_error());
        }
            echo "Record Successfully Added";

        closeconn();  

    }     

   }