跨月字段比较 2 个表之间的值 SQL Server
- 作者: Lzy8023hd
- 来源: 51数据库
- 2022-12-01
问题描述
当前演示:
命令表表达式返回是否曾订购过任何项目.然后,我们只需要计算/求和:
Current demo: http://sqlfiddle.com/#!18/7acdc/17
I am looking for a total of how many items were added in a specific month but then a total of how many were never ordered after that.
Tables:
CREATE TABLE Item (
ItemNo varchar(10)
,DateAdded varchar(10)
);
CREATE TABLE Order1 (
OrderNo int,
ItemNo varchar(10),
OrderDate varchar(10)
);
INSERT INTO Item (ItemNo, DateAdded)
VALUES ('111', 'Jan-17'),
('222', 'Jan-17'),
('333', 'Jan-17'),
('444', 'Feb-17'),
('555', 'Feb-17'),
('666', 'Feb-17');
INSERT INTO Order1 (ItemNo, OrderDate)
VALUES ('111', 'Jan-17'),
('111', 'Feb-17'),
('222', 'May-17'),
('333', 'Jan-17'),
('333', 'March-17'),
('444', 'Jan-17');
Currently i have:
SELECT
-- b.OrderDate,
A.DateAdded,
COUNT(DISTINCT A.ItemNo) AS [Items Added],
COUNT(CASE WHEN c.ItemNo IS NULL THEN 1 END) as [Items Never Ordered]
FROM Item a
CROSS JOIN (SELECT DISTINCT OrderDate FROM Order1) b
LEFT JOIN Order1 c
ON a.ItemNo = c.ItemNo
AND b.OrderDate = c.OrderDate
GROUP BY A.DateAdded
Which produces:
| DateAdded | Items Added | Items Never Ordered | |-----------|-------------|---------------------| | Feb-17 | 3 | 11 | | Jan-17 | 3 | 7 |
But i am looking for a result set such as:
| DateAdded | Items Added | Items Never Ordered | |-----------|-------------|---------------------| | Feb-17 | 3 | 2 | | Jan-17 | 3 | 0 |
I am struggling to get this working. Do i need a sub-query or something to match the items individually. Can anyone push me in the right direction? Thanks
解决方案
Maybe this:
WITH DataSource AS
(
SELECT A.*
,MAX([IsEverOrdered]) OVER (PARTITION BY ItemNo) [IsEverOrdered]
FROM Item A
OUTER APPLY
(
SELECT 1 [IsEverOrdered]
FROM Order1 B
WHERE A.[ItemNo] = B.[ItemNo]
) DS
)
SELECT DateAdded
,COUNT(DISTINCT ItemNo) AS [Items Added]
,SUM(CASE WHEN [IsEverOrdered] IS NULL THEN 1 ELSE 0 END) AS [Items Never Ordered]
FROM DataSource
GROUP BY DateAdded;
The Commant Table Expression returns if any item was ever ordered. Then, we just need to count/sum:
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