按值分组的列中连续月份的查询结果

Question

以下是示例数据：

Name | Hours | RDate       | Company |
------------------------------------
A    |0      |2014-08-01   |W
A    |0      |2014-07-01   |W
A    |0      |2014-06-01   |W
A    |0      |2014-05-01   |W
B    |0      |2014-08-01   |X
C    |0      |2014-07-01   |Y
C    |0      |2014-06-01   |Y
D    |0      |2014-08-01   |V     
D    |0      |2014-07-01   |Z

以下是我想要的结果：

Name | Hours | RDate       | Company |
------------------------------------
A    |0      |2014-08-01   |W
A    |0      |2014-07-01   |W
A    |0      |2014-06-01   |W
A    |0      |2014-05-01   |W
C    |0      |2014-07-01   |Y
C    |0      |2014-06-01   |Y

所以问题是：

如何只得到 RDate 列中连续月份的结果，即 2014-08-01, 2014-07-01（2014-08-01, 2014-06-01 不满足）相同的 name 和同样的company

Answer 1

我认为这是 Grouping Islands of Contiguous Dates 问题的变体。

;WITH Cte AS(
    SELECT *,
        RN = DATEADD(MONTH, - ROW_NUMBER() OVER (PARTITION BY Name, Company ORDER BY RDate), RDate)
    FROM Test
)
,CteCount AS(
    SELECT *,
        CC = COUNT(*) OVER(PARTITION BY Name, Company, RN)
    FROM Cte
)
SELECT
    Name, Hours, RDate, Company
FROM CteCount
WHERE CC > 1

SQL FIDDLE

Answer 2

虽然@wewesthemenace 的回答效率更高，但我尝试用我正在研究的解决方案弄清楚自己，并且它有效；将先前标记的答案保留为标记的原因会更好。这实际上也有效：

SELECT 
    one.*
FROM 
    foo one 
INNER JOIN 
    foo two
ON 
    (one.Name = two.Name and one.Company = two.Company) 
WHERE
    CONVERT(int,FORMAT(two.Date, 'yyyyMM')) - CONVERT(int,FORMAT(one.ACSS_Date, 'yyyyMM')) = 1
ORDER BY
    one.Name
    ,one.Date DESC

按值分组的列中连续月份的查询结果

Query Results For Consecutive Months In Column Grouped By Value

sql

sql-server

group-by