4

我正在尝试创建一个查询，该查询将从我正在创建的计费系统的四个表中获取信息。我有以下表格：

表发票

InvoiceID (PK)
ClientID
Date
Status
...

表客户端

ClientID (PK)
ClientName
...

表发票项

ItemID (PK)
InvoiceID
Amount
...

餐桌付款

PaymentID (PK)
InvoiceID
Amount
...

我需要创建一个查询，我可以在其中访问 Invoice 表中的信息以及客户名称，以及与发票关联的所有发票项目和付款的总和。

我尝试了以下方法：

SELECT 
    Invoice.InvoiceID, 
    Invoice.`Date`, 
    Invoice.Terms, 
    Invoice.DateDue, 
    Invoice.Status, 
    Client.ClinicName, 
    SUM(InvoiceItem.Amount), 
    SUM(Payment.PaymentAmount)
FROM Invoice
JOIN (Client, InvoiceItem, Payment) ON
    (Client.ClientID=Invoice.ClientID AND
     InvoiceItem.InvoiceID=Invoice.InvoiceID AND 
     Payment.InvoiceID=Invoice.InvoiceID)

虽然这种方法有效，但它将 SUM() 乘以用于获得总和的记录数（即，如果有两次付款 - 800,400 - 它给了我 (800+400)*2 -- 2400）。我猜我使用联接的方式有些问题，老实说，我从来不需要对多个表使用联接，而且我总是使用 GROUP BY，但我似乎无法让它正常工作.

更糟糕的是，过去几年我一直迷失在 vb.net/MSSQL 客户端编程的世界中，所以我的 MySQL 相当粗糙。

4

4 回答 4

6

您的问题是您不能在单个查询中一次聚合两个独立的表。但是，您可以使用子查询来做到这一点。

SELECT Invoice.InvoiceID, Invoice.`Date`, Invoice.Terms, Invoice.DateDue, Invoice.Status, Client.ClinicName, InvoiceItemSum.SumOfAmount, PaymentSum.SumOfPaymentAmount
  FROM Invoice
  INNER JOIN Client ON Client.ClientID = Invoice.ClientID
  INNER JOIN (
    SELECT InvoiceID, SUM(Amount) AS SumOfAmount
      FROM InvoiceItem
      GROUP BY InvoiceID
  ) InvoiceItemSum ON InvoiceItemSum.InvoiceID = Invoice.InvoiceID
  INNER JOIN (
    SELECT InvoiceID, SUM(PaymentAmount) AS SumOfPaymentAmount
    FROM Payment
    GROUP BY InvoiceID
  ) PaymentSum ON PaymentSum.InvoiceID = Invoice.InvoiceID
于 2012-08-31T00:02:02.113 回答
0

这里试试这个

SELECT  a.InvoiceID,
        a.`Date`,
        a.Terms,
        a.DateDue,
        a.Status,
        b.ClinicName,
        SUM(c.Amount),
        SUM(d.PaymentAmount)
FROM    Invoice a
            INNER JOIN Client b
                on a.ClientID = b.ClientID
            INNER JOIN InvoiceItem c
                ON c.InvoiceID = a.InvoiceID
            INNER JOIN JOIN Payment d
                ON d.InvoiceID = a.InvoiceID
GROUP BY    a.InvoiceID,
            a.`Date`,
            a.Terms,
            a.DateDue,
            a.Status,
            b.ClinicName

你能详细说明一下吗？

它将 SUM() 乘以用于获得总和的记录数（即，如果有两次付款 - 800,400 - 它给了我 (800+400)*2 -- 2400）

于 2012-08-31T00:01:28.817 回答
0

尝试这个：

SELECT
    Invoice.InvoiceID,
    Invoice.`Date`,
    Invoice.Terms,
    Invoice.DateDue,
    Invoice.Status,
    Client.ClinicName,
    SUM(InvoiceItem.Amount),
    SUM(Payment.PaymentAmount)
FROM Invoice 
JOIN Client ON Client.ClientID=Invoice.ClientID
JOIN InvoiceItem ON InvoiceItem.InvoiceID=Invoice.InvoiceID
JOIN Payment ON Payment.InvoiceID=Invoice.InvoiceID
group by 1,2,3,4,5,6;

我对您的查询做了两件事：

  1. 为每个子表创建单独的连接
  2. 添加了 a group by，否则总和将无法正常工作（仅供参考，在所有其他数据库中，省略 group by 实际上会导致语法错误）
于 2012-08-30T23:58:28.137 回答
0

你也可以通过“CROSS APPLY”来实现它

SELECT Invoice.InvoiceID, Invoice.`Date`, Invoice.Terms, Invoice.DateDue, Invoice.Status, Client.ClinicName, InvoiceItemSum.SumOfAmount, PaymentSum.SumOfPaymentAmount
  FROM Invoice
  INNER JOIN Client ON Client.ClientID = Invoice.ClientID
  CROSS APPLY ( SELECT ISNULL(SUM(Amount),0) AS SumOfAmount 
                                  FROM   InvoiceItem 
                                  WHERE  InvoiceID = Invoice.InvoiceID
                                ) InvoiceItemSum 
  CROSS APPLY ( SELECT ISNULL(SUM(PaymentAmount),0) AS SumOfPaymentAmount 
                                  FROM   Payment 
                                  WHERE  InvoiceID = Invoice.InvoiceID
                                ) PaymentSum
于 2016-05-25T13:59:14.687 回答