获得付款总额减去收到的金额答案

【问题标题】：Get payment total minus amount received获得付款总额减去收到的金额
【发布时间】：2014-11-12 09:50:42
【问题描述】：

我有两张桌子。

1.发票

invoice_Id  invoice_no  client_id   date        total_price    
-----------------------------------------------------------------------------
2           INV00001    9           2014-10-15  200.00   
7           INV00002    9           2014-10-16  560.00   
8           INV00003    9           2014-10-21  100.00   
11          INV00004    9           2014-10-27  101.00

2.Invoice_payment

InvPayment_id   client_id   Invoice_Id  receipt_no  payment_date    amount_received discount
--------------------------------------------------------------------------------------------
6               9           8           REC00002    2014-10-31      5.00            0.00

现在我想通过合计发票金额并减去收到的任何金额来获得客户应付的总金额。

预期结果：

client_id   Total_price    Due_Amount
-----------------------------------------------------------------------------
9           961.00         956.00

注意事项：

如果迄今为止没有付款，则行数为零。
如果多笔付款，可能会有多行。

这是我尝试过的：

;WITH cte (clientid, invoiceid,  paid, disc)
As
(
    Select client_id clientId, invoice_Id invoiceId,  sum(amount_received) paid, sum(discount) disc
    From tbl_Invoice_Payment
    Group by invoice_id, client_id
)
Select I.client_id, invoice_Id, invoice_no, I.due_date
,SUM(I.total_price), Isnull(SUM(paid), 0) Paid, (SUM(Total_price) - Isnull(SUM(paid),0) - Isnull(SUM(disc),0)) Balance
--,I.total_price, Isnull(paid, 0) Paid, (Total_price - Isnull(paid,0) - Isnull(disc,0)) Balance
From tbl_invoice I Left join cte On I.client_id = cte.clientId 
        And I.invoice_id = cte.invoiceid
        left join tbl_client C on C.client_id = I.client_id
group by I.client_id, invoice_Id, invoice_no, due_date, account_type, company_name, total_price, paid, disc
order by company_name

但它没有按预期工作。

【问题讨论】：

标签： sql-server tsql group-by common-table-expression

【解决方案1】：

您可以简单地添加一个子查询来返回付款结果并从总数中减去该值，而不是使用 CTE：

SQL Fiddle Demo

架构设置：

CREATE TABLE Invoice
    ([invoice_Id] int, [invoice_no] varchar(8), [client_id] int, 
     [date] datetime, [total_price] decimal(5,2));

INSERT INTO Invoice
    ([invoice_Id], [invoice_no], [client_id], [date], [total_price])
VALUES
    (2, 'INV00001', 9, '2014-10-15 00:00:00', 200.00),
    (7, 'INV00002', 9, '2014-10-16 00:00:00', 560.00),
    (8, 'INV00003', 9, '2014-10-21 00:00:00', 100.00),
    (11, 'INV00004', 9, '2014-10-27 00:00:00', 101.00);

CREATE TABLE Invoice_Payment
    ([InvPayment_id] int, [client_id] int, [Invoice_Id] int, [receipt_no] varchar(8), 
     [payment_date] datetime, [amount_received] decimal(5,2), [discount] int);

INSERT INTO Invoice_Payment
    ([InvPayment_id], [client_id], [Invoice_Id], [receipt_no], [payment_date], 
     [amount_received], [discount])
VALUES
    (6, 9, 8, 'REC00002', '2014-10-31 00:00:00', 5.00, 0.00);

查询生成输出：：

SELECT  i.client_id , SUM(i.total_price) AS Total_price,
        SUM(i.total_price) - ( SELECT   SUM(ip.amount_received)
                               FROM     dbo.Invoice_Payment ip
                               WHERE    i.client_id = ip.client_id
                             ) AS DueAmount
FROM    dbo.Invoice i
WHERE   client_id = 9
GROUP BY client_id

Results：

| CLIENT_ID | TOTAL_PRICE | DUEAMOUNT |
|-----------|-------------|-----------|
|         9 |         961 |       956 |

【讨论】：

太棒了。查询中有一些修改，但我已经让它工作了。谢谢

【解决方案2】：

请选择您想要的值。问题与分组有关。

在group by 子句中，I.client_id、invoice_Id、invoice_no、due_date、account_type、company_name、total_price、paid、disc 指定了这么多字段。 invoice_Id, invoice_no 阻止您需要的分组。从select 和group by 中删除invoice_Id 和invoice_no，然后重试。

【讨论】：