SQL连接,按三个表分组以获取总数

我不确定我抓到你了,但这可能是你要找的:

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,这将为您获得金额总和

非常感谢您的回复!

Saggi Malachi,不幸的是,如果有不止一笔付款,这个查询将汇总发票金额。假设一张39美元的18美元和12美元的发票有两次付款。因此,与其最终得到一个看起来像是:

1   39.00   9.00

你最终会得到:

1   78.00   48.00

Charles Bretana,在将我的查询精简为尽可能简单的查询的过程中,我(愚蠢地)省略了一个额外的表customerinvoices,它提供了客户和发票之间的链接。这可用于查看尚未付款的发票。

经过许多努力,我认为下面的查询将返回我需要的内容:

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?

对于那些希望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取点表。所以它们之间的连接是1:N(一个用户,多个点记录)。

现在,在“points”表中,我还存储了有关用户获得点数的信息(登录、单击横幅等)。我想列出 SUM(points) 然后 SUM(points WHERE type = x) . 也就是说,按用户拥有的所有点数排序,然后按用户为特定操作(例如登录)获得的点数排序。

SQL将是:

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

它的美丽在于 SUM(points.points * (points.type = 7)) 其中,内圆括号的计算结果为0或1,从而将给定的点值乘以0或1,具体取决于它是否等于所需的点类型。

首先,发票表中不应该有CustomerId吗?实际上,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,则该发票甚至不会显示在查询的输出中,即使它是外部连接。。。

另外,当客户付款时,你如何知道要附在哪张发票上?如果唯一的方法是通过付款存根上的InvoiceId,那么您是(也许是不恰当地)将发票与付款的客户关联起来,而不是与订购它们的客户关联起来。。。(有时发票可以由订购服务的客户以外的人支付)

我知道这很晚了,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.
*/