对三个表进行 SQL JOIN、GROUP BY 以获得总计

Question

我继承了以下数据库设计。 表如下：

customers
---------
customerid  
customernumber

invoices
--------
invoiceid  
amount

invoicepayments
---------------
invoicepaymentid  
invoiceid  
paymentid

payments
--------
paymentid  
customerid  
amount

我的查询需要返回给定客户编号的发票 ID、发票金额（在发票表中）和应付金额（发票金额减去已向发票支付的所有付款）。 一个客户可能有多张发票。

当对一张发票进行多次付款时，以下查询会给出重复的记录：

SELECT i.invoiceid, i.amount, i.amount - p.amount AS amountdue
FROM invoices i
LEFT JOIN invoicepayments ip ON i.invoiceid = ip.invoiceid
LEFT JOIN payments p ON ip.paymentid = p.paymentid
LEFT JOIN customers c ON p.customerid = c.customerid
WHERE c.customernumber = '100'

我该如何解决这个问题？

Answer 1

我不确定我是否明白你的意思，但这可能就是你正在寻找的：

SELECT i.invoiceid, sum(case when i.amount is not null then i.amount else 0 end), sum(case when i.amount is not null then i.amount else 0 end) - sum(case when p.amount is not null then p.amount else 0 end) AS amountdue
FROM invoices i
LEFT JOIN invoicepayments ip ON i.invoiceid = ip.invoiceid
LEFT JOIN payments p ON ip.paymentid = p.paymentid
LEFT JOIN customers c ON p.customerid = c.customerid
WHERE c.customernumber = '100'
GROUP BY i.invoiceid

如果每张发票有多个付款行，这将为你提供金额

Answer 2

非常感谢您的回复！

Saggi Malachi，不幸的是，在有不止一笔付款的情况下，该查询会汇总发票金额。 假设一张 39 美元的发票有两笔付款：18 美元和 12 美元。 因此，与其最终得到如下结果：

1   39.00   9.00

你最终会得到：

1   78.00   48.00

Charles Bretana，在将查询精简为最简单的查询的过程中，我（愚蠢地）省略了一个附加表 customerinvoices，它提供了一个链接客户和发票之间。 这可用于查看尚未付款的发票。

经过一番努力，我认为以下查询会返回我需要的内容：

SELECT DISTINCT i.invoiceid, i.amount, ISNULL(i.amount - p.amount, i.amount) AS amountdue
FROM invoices i
LEFT JOIN invoicepayments ip ON i.invoiceid = ip.invoiceid
LEFT JOIN customerinvoices ci ON i.invoiceid = ci.invoiceid
LEFT JOIN (
  SELECT invoiceid, SUM(p.amount) amount
  FROM invoicepayments ip 
  LEFT JOIN payments p ON ip.paymentid = p.paymentid
  GROUP BY ip.invoiceid
) p
ON p.invoiceid = ip.invoiceid
LEFT JOIN payments p2 ON ip.paymentid = p2.paymentid
LEFT JOIN customers c ON ci.customerid = c.customerid
WHERE c.customernumber='100'

你们同意吗？

Answer 3

对于那些想要从同一个表中获取各种聚合值的人，我有一个提示。

假设我有用户表和用户获得的积分表。 所以它们之间的联系是1:N（一个用户，很多点记录）。

现在，在“积分”表中，我还存储有关用户获得积分的信息（登录、单击横幅等）。 我想列出按 SUM(points) 排序的所有用户，然后按 SUM(points WHERE type = x) 排序。 也就是说，按用户拥有的所有积分排序，然后按用户针对特定操作（例如登录）获得的积分排序。

SQL 如下：

SELECT SUM(points.points) AS points_all, SUM(points.points * (points.type = 7)) AS points_login
FROM user
LEFT JOIN points ON user.id = points.user_id
GROUP BY user.id

其美妙之处在于 SUM(points.points * (points.type = 7))，其中内括号的计算结果为 0 或 1，从而将给定的点值乘以0 或 1，取决于它是否等于我们想要的点类型。

Answer 4

首先，Invoices表中不应该有一个CustomerId吗？ 事实上，您无法对尚未付款的发票执行此查询。 如果发票上没有付款，该发票甚至不会显示在查询的输出中，即使它是外部联接...

另外，当客户付款时，您如何知道要附加什么发票到 ？ 如果唯一的方法是通过随付款一起到达的存根上的 InvoiceId，那么您（可能不恰当）将发票与付款的客户关联，而不是与订购发票的客户关联。 （有时发票可以由订购服务的客户以外的其他人支付）

Answer 5

我知道这已经晚了，但它确实回答了你原来的问题。

/*Read the comments the same way that SQL runs the query
    1) FROM 
    2) GROUP 
    3) SELECT 
    4) My final notes at the bottom 
*/
SELECT 
        list.invoiceid
    ,   cust.customernumber 
    ,   MAX(list.inv_amount) AS invoice_amount/* we select the max because it will be the same for each payment to that invoice (presumably invoice amounts do not vary based on payment) */
    ,   MAX(list.inv_amount) - SUM(list.pay_amount)  AS [amount_due]
FROM 
Customers AS cust 
    INNER JOIN 
Payments  AS pay 
    ON 
        pay.customerid = cust.customerid
INNER JOIN  (   /* generate a list of payment_ids, their amounts, and the totals of the invoices they billed to*/
    SELECT 
            inpay.paymentid AS paymentid
        ,   inv.invoiceid AS invoiceid 
        ,   inv.amount  AS inv_amount 
        ,   pay.amount AS pay_amount 
    FROM 
    InvoicePayments AS inpay
        INNER JOIN 
    Invoices AS inv 
        ON  inv.invoiceid = inpay.invoiceid 
        INNER JOIN 
    Payments AS pay 
        ON pay.paymentid = inpay.paymentid
    )  AS list
ON 
    list.paymentid = pay.paymentid
    /* so at this point my result set would look like: 
    -- All my customers (crossed by) every paymentid they are associated to (I'll call this A)
    -- Every invoice payment and its association to: its own ammount, the total invoice ammount, its own paymentid (what I call list) 
    -- Filter out all records in A that do not have a paymentid matching in (list)
     -- we filter the result because there may be payments that did not go towards invoices!
 */
GROUP BY
    /* we want a record line for each customer and invoice ( or basically each invoice but i believe this makes more sense logically */ 
        cust.customernumber 
    ,   list.invoiceid 
/*
    -- we can improve this query by only hitting the Payments table once by moving it inside of our list subquery, 
    -- but this is what made sense to me when I was planning. 
    -- Hopefully it makes it clearer how the thought process works to leave it in there
    -- as several people have already pointed out, the data structure of the DB prevents us from looking at customers with invoices that have no payments towards them.
*/