Question

嘿伙计们我是OOP的新手，我想要运行两个需要相同连接的功能。我正在努力以一种允许两个函数使用它的方式声明$ mysqli变量。我宁愿只使用一个，因为我稍后会添加更多功能。非常感谢任何帮助。

我收到的错误信息是;

注意：未定义的变量：第13行的C：\ Users \ PC \ Documents \ XAMPP \ htdocs \ test.php中的mysqli

<?php
class OOP 
{
function __construct(){
$mysqli = new mysqli('localhost', 'c3337015', 'c3337015', 'iitb');
if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
        }
    }

function getMember(){
        $stmt = $mysqli->prepare("SELECT First_Name FROM forum members");
        $stmt->execute();
        $stmt->store_result();
        $stmt->bind_result($First_Name);

        while ($row = $stmt->fetch()) {
        echo $First_Name;
        }

}
function getForum(){
        $stmt = $mysqli->prepare("SELECT ForumTitle FROM forum");
        $stmt->execute();
        $stmt->store_result();
        $stmt->bind_result($ForumTitle);

        while ($row = $stmt->fetch()) {
        echo $ForumTitle;
        }

}

}

Answer 1

您在构造函数中声明$mysqli，但它不是类变量。您只需将其添加为类变量：

class OOP {
    private $mysqli;
    ...

然后，只要您想要访问该变量，请将$mysqli替换为$this->mysqli。

Answer 2

<?php
/**
 * My file
 * @author yourname
 *
 */
class OOP {

    /**  ============ class vars  ============ **/

    /**
     * DocBloc comment
     * @var mysqli
     */
    private $mysqli = null;



    /** ============ class methods ============ **/

    function __construct() 
    {

        $this->mysqli = new mysqli ( 'localhost', 'c3337015', 'c3337015', 'iitb' );

        if (mysqli_connect_errno ()) {
            printf ( "Connect failed: %s\n", mysqli_connect_error () );
            exit ();
        }
    }


    /*
     * DocBloc
     */
    function getMember() 
    {
        $stmt = $this->mysqli->prepare ( "SELECT First_Name FROM forum members" );
        $stmt->execute ();
        $stmt->store_result ();
        $stmt->bind_result ( $First_Name );

        while ( $row = $stmt->fetch () ) {
            echo $First_Name;
        }
    }


    /*
     * DocBloc
     */
    function getForum() 
    {
        $stmt = $this->mysqli->prepare ( "SELECT ForumTitle FROM forum" );
        $stmt->execute ();
        $stmt->store_result ();
        $stmt->bind_result ( $ForumTitle );

        while ( $row = $stmt->fetch () ) {
            echo $ForumTitle;
        }
    }
}

PHP中的OOP - “__construct”连接

2 个答案: