Question

我正在制作一个电子商务网络应用程序，其中包含佣金作为功能之一，我正在努力。

订单表：

CREATE TABLE `orders` (
 `OrderId` int(11) NOT NULL AUTO_INCREMENT,
 `OrderCode` varchar(20) NOT NULL,
 `CustEmailAdd` varchar(80) NOT NULL,
 `CustDelAddId` varchar(255) NOT NULL,
 `ProdCode` varchar(255) NOT NULL,
 `Quantity` varchar(100) NOT NULL,
 `PaytMethod` varchar(255) NOT NULL,
 `ShippingCharges` float NOT NULL,
 `TaxedAmount` float NOT NULL,
 `AppliedCredits` int(11) NOT NULL,
 `PayableAmount` varchar(255) NOT NULL,
 `OrderDate` datetime NOT NULL,
 `OrderModified` datetime NOT NULL,
 `OrderStatus` varchar(255) NOT NULL,
 `OrderIPAddress` varchar(20) NOT NULL,
 PRIMARY KEY (`OrderId`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8

佣金表：

CREATE TABLE `commission` (
 `CommId` int(11) NOT NULL AUTO_INCREMENT,
 `OrderCode` varchar(20) NOT NULL,
 `ProdCode` varchar(255) NOT NULL,
 `ModCode` varchar(255) NOT NULL,
 `AffiCode` varchar(255) NOT NULL,
 `BenCode` varchar(255) NOT NULL,
 `ModCommAmount` varchar(255) NOT NULL,
 `AffiCommAmount` varchar(255) NOT NULL,
 `BenCommAmount` varchar(255) NOT NULL,
 PRIMARY KEY (`CommId`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8

affiliate_commission_payable表

CREATE TABLE `affiliate_commission_payable` (
 `ACP_Id` int(11) NOT NULL AUTO_INCREMENT,
 `CommId` int(11) NOT NULL,
 `ACP_PaymentStatus` varchar(255) NOT NULL,
 `ACP_PaymentDate` varchar(255) NOT NULL,
 `ACP_PaymentDetails` text NOT NULL,
 PRIMARY KEY (`ACP_Id`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8

我有一个像这样的数组：

// For Order 000001    
    Array
    (
        [0] => ORD-000001
    )
    Array
    (
        [Paid] => 26.25
        [Due] => 42.75
    )

// For Order 000002
    Array
    (
        [0] => ORD-000002
    )
    Array
    (
        [Cancelled] => 33.75
        [Due] => 13.5
    )
// For Order 000003
    Array
    (
        [0] => ORD-000003
    )
    Array
    (
        [Paid] => 13.50
        [Paid] => 14.25
    )

我要做的是向用户显示订单佣金摘要。它的表格格式如下：

OrderCode    Paid Comm    Cancelled Comm    Due Comm    View Details

ORD-000001   addAllPaid   addAllCancelled   addAllDue       View

ORD-000002   addAllPaid   addAllCancelled   addAllDue       View

ORD-000003   addAllPaid   addAllCancelled   addAllDue       View

如果数组中没有值用于委托，请将其替换为0，否则添加它们并根据订单代码显示它。我如何实现这一目标？

到目前为止我尝试过的代码： - ＆gt;

<?php
$qu = "SELECT
           c.*, o.*, acp.*, .*, a.*
       FROM
           customers c, orders o, affiliate_commission_payable acp, commission com, affiliates a
       WHERE
           a.AffiCode = '".$affiCode."'
               AND
           c.AffiCode = '".$affiCode."'
               AND
           c.CustEmailAdd = o.CustEmailAdd
               AND
           acp.CommId = com.commId
               AND
           com.OrderCode = o.OrderCode
               AND
           com.AffiCode = c.AffiCode";
$validate->Query($qu);
if ($validate->NumRows() >= 1) {
    while ($rows = $validate->FetchAllDatas()) {
        $commId = $rows["CommId"];
        $orderCode = $rows["OrderCode"];
        $orderDate = $rows["OrderDate"];
        $arrCommissionStatus = explode(', ', $rows["ACP_PaymentStatus"]);
        $prdCodes = explode(', ', $rows["ProdCode"] );
        $arrAffiCommAmount = explode(', ', $rows["AffiCommAmount"]);

        $ord = explode(', ', $rows["OrderCode"]);

        $prdCodeAndCommStatus = array_combine($prdCodes, $arrCommissionStatus);
        $arrCommAndStatus = array_combine($arrCommissionStatus, $arrAffiCommAmount);

        $table .= "<tr>";
        $table .= "<td>".$orderCode."</td>";
        $table .= "<td>".$orderDate."</td>";
        $table .= "<td>".count($prdCodes)."</td>";

        foreach ($arrCommAndStatus as $key => $value) {
            if ($value == 'Due') {
                $totDueComm += $key;
            } else {
                $totDueComm = 0;
            }
        }
        foreach ($arrCommAndStatus as $key => $value) {
            if ($value == 'Cancelled') {
                $totCancellComm += $key;
            } else {
                $totCancellComm = 0;
            }
        }
        foreach ($arrCommAndStatus as $key => $value) {
            if ($value == 'Paid') {
                $totPaidComm += $key;
            } else {
                $totPaidComm = 0;
            }
        }

        $table .= "<td>".number_format($totPaidComm, 2)."</td>";
        $table .= "<td>".number_format($totCancellComm, 2)."</td>";
        $table .= "<td>".number_format($totDueComm, 2)."</td>";
        $table .= "<td>".number_format(array_sum($arrAffiCommAmount), 2)."</td>";
        $table .= "<td><a href='//www.example.com/nbs/Affiliates/Commission.php?ord=".$orderCode."&id=".$commId."'>View</a></td>";
        $table .= "</tr>";

    }   
}

更新1 ：

我以某种方式试图完成它。这里是我必须在上面的while循环中更改的代码：

$strQ = "SELECT com.*, acp.* FROM commission com, affiliate_commission_payable acp WHERE com.AffiCode = '".$affiCode."' AND com.OrderCode = '".$orderCode."' AND com.CommId = acp.CommId";
$validate->Query($strQ);
if ($validate->NumRows() >= 1) {
    while ($rows_strQ = $validate->FetchAllDatas()) {
        $comAmt = explode(', ', $rows_strQ["AffiCommAmount"]);
        $coStat = explode(', ', $rows_strQ["ACP_PaymentStatus"]);
        $ts = array_combine( $comAmt, $coStat);
        foreach ($ts as $key => $value) {
            if (in_array($value, $sta) && $value == 'Due') {
                $totDueComm += $key;
                $totalDueCommission += $totDueComm;
            }                       
        }
    }
}

但是更新刚刚完成了第一个订单而不是所有订单。我哪里出错了？请帮助我。感谢。

更新2 ：

我以某种方式管理它以完成它..但是有一个问题是金额被添加到先前的订单金额然后显示金额..例如，说我有Rs。 300 in Order 1和卢比。 200 in Order 2 ..因此，对于Order 1，它会显示Rs. 300，而对于Order 2，显示的金额为Rs. 500。我希望它只显示Rs. 200而不是500卢比。我该如何解决这个问题？请帮帮我..再次感谢。

这是我到目前为止使用的代码：

$sta = array('Due', 'Cancelled', 'Paid');

$qu = "SELECT c.*, o.*, acp.*, com.*, a.* FROM customers c, orders o, affiliate_commission_payable acp, commission com, affiliates a WHERE a.AffiCode = '".$affiCode."' AND c.AffiCode = '".$affiCode."' AND c.CustEmailAdd = o.CustEmailAdd AND acp.CommId = com.commId AND com.OrderCode = o.OrderCode AND com.AffiCode = c.AffiCode GROUP BY o.OrderCode";
$validate->Query($qu);
if ($validate->NumRows() >= 1) {
    while ($rows = $validate->FetchAllDatas()) {
        $commId = $rows["CommId"];
        $orderCode = $rows["OrderCode"];
        $orderDate = $rows["OrderDate"];
        $arrCommissionStatus = explode(', ', $rows["ACP_PaymentStatus"]);
        $prdCodes = explode(', ', $rows["ProdCode"] );
        $arrAffiCommAmount = explode(', ', $rows["AffiCommAmount"]);

        $prdCodeAndCommStatus = array_combine($prdCodes, $arrCommissionStatus);
        $arrCommAndStatus = array_combine($arrAffiCommAmount, $arrCommissionStatus);

        $table .= "<tr>";
        $table .= "<td>".$orderCode."</td>";
        $table .= "<td>".$orderDate."</td>";
        $table .= "<td>".count($prdCodes)."</td>";

        foreach ($rows as $key => $value) {
            if ($key === "AffiCommAmount") {
                $comAmt = explode(', ', $rows["AffiCommAmount"]);
            }
            if ($key === "ACP_PaymentStatus") {
                $coStat = explode(', ', $rows["ACP_PaymentStatus"]);
            }
        }
        $ts = array_combine( $comAmt, $coStat);

        foreach ($ts as $k => $v) {
            if (in_array($v, $sta) && $v == 'Due') {
                $totDueComm += $k;
                $totalDueCommission = $totDueComm;
            }
        }



        $table .= "<td>".number_format($totPaidComm, 2)."</td>";
        $table .= "<td>".number_format($totCancellComm, 2)."</td>";
        $table .= "<td>".number_format($totDueComm, 2)."</td>";
        $table .= "<td>".number_format(array_sum($arrAffiCommAmount), 2)."</td>";
        $table .= "<td><a href='//www.example.com/nbs/Affiliates/Commission.php?ord=".$orderCode."&id=".$commId."'>View</a></td>";
        $table .= "</tr>";
    }

}

？＆GT;

Answer 1

根据聊天情况，这里有解决方案：http://ideone.com/tXoWDk

我不确定是否了解数据库结构中的所有字段，但如果如您所说，想要汇总数据库中每个订单的佣金，请参阅以下示例：

select 
    o.code, sum(c.amount) as total
from orders o
inner join com c on c.orderId = o.id
group by o.code

关键是在PHP中这样做是没用的，因为MySQL也可以处理它而无需进一步处理，所以如果我简化这样的结构：

mysql> select * from orders;
+----+------------+
| id | code       |
+----+------------+
|  1 | ORDER-0001 |
|  2 | ORDER-0002 |
+----+------------+

mysql> select * from com;
+----+--------+---------+-----------+
| id | amount | orderId | status    |
+----+--------+---------+-----------+
|  1 |     15 |       1 | DUE       |
|  2 |     12 |       1 | DUE       |
|  3 |      2 |       2 | CANCELLED |
|  4 |     23 |       2 | PAID      |
|  5 |      3 |       1 | CANCELLED |
|  6 |      5 |       1 | PAID      |
|  7 |     12 |       2 | CANCELLED |
+----+--------+---------+-----------+


mysql> select
    -> o.code, c.status, sum(c.amount) as total
    -> from orders o
    -> inner join com c on c.orderId = o.id
    -> group by c.status, o.code
    -> order by o.code, c.status, c.amount;
+------------+-----------+-------+
| code       | status    | total |
+------------+-----------+-------+
| ORDER-0001 | CANCELLED |     3 |
| ORDER-0001 | DUE       |    27 |
| ORDER-0001 | PAID      |     5 |
| ORDER-0002 | CANCELLED |    14 |
| ORDER-0002 | PAID      |    23 |
+------------+-----------+-------+

并总计总和，只需将以下内容应用于您的结果：

$ordersTotalCommissions = [];

// fetch your data
foreach ($result as $line) {

    // initialize the 'order' group by putting 0 as a commission
    if (!isset($ordersTotalCommissions[$line['code']])) {
        $ordersTotalCommissions[$line['code']] = 0;
    }

    // sum each commission
    $ordersTotalCommissions[$line['code']] += $line['total'];
}

print_r($ordersTotalCommissions);

告诉我这是否有助于你

无法添加数组值，有什么问题？

1 个答案: