douyamitong57935 2015-10-23 09:51
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我需要将mysqli连接对象返回给另一个类

<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>

when I'm trying to use the $ob which contains the mysqli connect object returned by connect class it is showing expecting mysqli parameter.

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1条回答 默认 最新

  • dtlygweb2017 2015-10-23 10:09
    关注

    use $ob = new connect(); and do not use mysqli_connect in constructer make a public function and like

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

    and then use this

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

    hope this works for you

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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