php错误发生警告:mysqli :: close():无法在php中获取mysqli错误

Question

我有问题要建立一个连接到DB的代码,然后关闭连接.

我的DB类如下:

class DataBase {
    private $conn;
    private $info;
    private $db;

    public function __construct ($servername=DB_SERVER, $user=DB_USER, $pass=DB_PASS, $db=DB_DATABASE) {
                $this->db = $db;
                $this->conn = new mysqli($servername,$user,$pass, $this->db);


                if (!$this->conn)
                {
                    if  ($this->conn->connect_error) {
                        $this->info = "ERROR CODE=DB100: Connection failed <br> " . $conn->connect_error;
                        $this->close();
                        echo $this->getInfo();  
                    }
                    else {
                        $this->info = "ERROR CODE=DB101: Connection failed <br> ";
                        $this->close();
                        echo $this->getInfo();
                    }
                }
    }

    public function getInfo () {
        return $this->info;
    }

    // send sql to database
    public function sendQuery ($sql, $numr=null, $start=null) {

        if ($numr) {
            if ($start) 
                $sql .= " LIMIT $start, $numr";
            else 
                $sql .= " LIMIT $numr";
        }

                    $this->conn->set_charset("utf8");
        $result = $this->conn->query($sql);

        if ($result == true) {
                        return $result;
        }
        else {
                        $this->info = "ERROR CODE=DB102: Sql query not work <br>".$this->conn->error;
                        $this->close();
                        echo $this->getInfo();
                        return false;
        }

    }

            public function sendQueryNoneCahrSet ($sql, $numr=null, $start=null) {
                    if ($numr) {
                        if ($start) 
                                $sql .= " LIMIT $start, $numr";
                        else 
                                $sql .= " LIMIT $numr";
        }

        $result = $this->conn->query($sql);

        if ($result == true) {
            return $result;
        }
        else {
            $this->info = "ERROR CODE=DB103: Sql query not work <br>".$this->conn->error;
            $this->close();
            echo $this->getInfo();
            return false;
        }
            }


    // send sql query safe to avoid sql injection.
    public function getConn () {
                return $this->conn;
    }

    public function getDataBase (){
                return $this->db;
            }


    // This method close the connection
    public function close () {
                $this->conn->close();
    }

    function __destruct() {
                $this->close();
    }
}

当我运行这段代码时:

           $con = @new DataBase();
           $r = $con->sendQuery($sql);
           $con->close();

我收到了这个错误:

Warning: mysqli::close(): Couldn't fetch mysqli in C:\Program Files (x86)\EasyPHP-DevServer-14.1VC11\data\localweb\projects\automasion\admin\protected\componets\database.php on line 102

程序正确运行并将数据设置为DB但发生此错误.

错误指向:

$this->conn->close();

在我的数据库连接代码中.

Answer 1

在构造函数中,您创建了一个mysqli实例:

$this->conn = new mysqli($servername,$user,$pass, $this->db);

但是$this->db 为null并且没有价值.你应该把它改成:

$this->conn = new mysqli($servername,$user,$pass,$db);

还有一件事:

function __destruct() {
                $this->close();
    }

这个函数在php想要销毁你的对象时运行,但你已经关闭了你的连接!

$con = @new DataBase();
$r = $con->sendQuery("show tables");
$con->close();