Question

<body>
    <?php
        if(isset($_GET["submit"])){
        class connect{
            var $connect;
            function __construct(){
                $connect=mysqli_connect("","root","","bank");
                return $connect;
            }
        }
        class deposit{
            var $ac,$am;
            function __construct($accno,$amount){
                $ac=$accno;
                $am=$amount;
                $ob = new connect;
                $result=mysqli_query($ob,"SELECT * FROM customer WHERE accno='$ac'"); //Warning: mysqli_query() expects parameter 1 to be mysqli, object given in C:\xampp\htdocs\bank\deposit.php on line 25
                $row=mysqli_fetch_array($result);
                $total=$am+$row['balance'];
                mysqli_query($ob,"UPDATE customer SET balance='$total' where accno='$ac'");
            }
        }
        $amount=$_GET["amount"];
        $accno=$_GET["accno"];
        $ob1 = new deposit($accno,$amount); 
        }
    ?>
</body>

当我尝试使用包含连接类返回的mysqli连接对象的$ob时，它显示了期望的mysqli参数。

Answer 1

使用$ob = new connect();并且不要在constructer中使用mysqli_connect创建公共函数并且喜欢

class connect{ var $connect; public function conn(){ $connect=mysqli_connect("","root","","bank"); return $connect; }

然后使用此

mysqli_query($ob->conn(),"UPDATE customer SET balance='$total' where accno='$ac'");

希望这对你有用

我需要将mysqli连接对象返回给另一个类

1 个答案: