参见英文答案 > Call to a member function on a non-object                                     8个
我是新手以OOP方式使用PHP但发现我的数据库连接类有问题.

这里有一个MysqLi连接类的文件

$db_name = 'dbname';
$db_user = 'dbuser';        
$db_password = 'dbpassword';
$db_host = 'localhost';

class database {

    public $MysqLi;

    public function connect($db_host,$db_user,$db_password,$db_name){

        $this->MysqLi = new MysqLi($db_host,$db_name);

        if ($MysqLi->connect_errno) {
            return "Sorry Andre,but you seem to have messed up the DB connection :(";
        }
    }
}

$newConnection = new database;
$newConnection->connect($db_host,$db_name);

然后我想在另一个文件中的数据库连接中使用变量$MysqLi – 这是使用$MysqLi变量连接到数据库的简单插入.我将上面的内容包含在连接文件中,但是当我在数据库类中调用该方法时,似乎没有返回$MysqLi变量.我得到PHP错误说…

Fatal error: Call to a member function PRepare() on a non-object in...

我见过那个用

global $MysqLi;

然而,我想以正确的方式去做,因为我听说这不是好的做法.

我知道我可能在这里做错了,因为我不熟悉使用OOP但是我假设通过在connect函数中返回该变量,我可以通过在外部创建类来访问它.

感谢帮助,
谢谢.

$newConnection = new database;
$newConnection->connect($db_host,$db_name);
$newConnection->MysqLi /* here you have access From outside */

// like this from inside
if ($this->MysqLi->connect_errno) {
    return "Sorry Andre,but you seem to have messed up the DB connection :(";
}

private $MysqLi; 
  // instead of 
  public $MysqLi;