Question

好的，我有一份月份订单清单，这样就很容易了。

SELECT COUNT(*) AS ITEMS,

下一部分易于：

COUNT(DISTINCT(PICKSET_NO))AS PICKSETS,

下一部分我无法解决：

SUM(ITEMS/PICKSETS) AS AVGPICKSETSIZE
FROM dbo.orders

感谢您的帮助。这是一个块中的代码。

SELECT
COUNT(*) AS ITEMS,
COUNT(DISTINCT(PICKSET_NO))AS PICKSETS,
SUM(ITEMS/PICKSETS)
FROM dbo.CollationOrders
GO

Answer 1

重复表达式：

SELECT COUNT(*) AS ITEMS,
       COUNT(DISTINCT PICKSET_NO) AS PICKSETS,
       COUNT(*) / (1.0 * COUNT(DISTINCT PICKSET_NO))
FROM dbo.CollationOrders;

您无法在同一select中重复使用列别名。

1.0是为了防止整数除法。

Answer 2

您可以这样做：

SELECT ITEMS,
       PICKSETS,
       ITEMS / (1.0 * PICKSETS)
FROM (
       SELECT COUNT(*) AS ITEMS,
              COUNT(DISTINCT PICKSET_NO) AS PICKSETS     
       FROM TableName)t

Answer 3

我最终使用了：

SELECT
    COUNT(*) AS ITEMS,
    COUNT(DISTINCT om.PicksetNo) AS PICKSETS ,
    FORMAT(COUNT(*) / (1.0 * COUNT(DISTINCT om.PicksetNo)),'N2'),
   ca.YWK AS YWK
FROM CHDS_Common.dbo.OMOrder om
INNER JOIN CHDS_Management.dbo.Calendar ca ON om.EarliestPickDate = ca.DT
GROUP BY ca.YWK

将COUNT列除以COUNT（DISTINCT（列给予平均订单大小SQL SERVER

3 个答案: