SQL计数连续几天

这是SQL数据库数据：

UserTable

UserName    | UserDate      | UserCode
-------------------------------------------
user1       | 08-31-2014    | 232
user1       | 09-01-2014    | 232
user1       | 09-02-2014    | 0
user1       | 09-03-2014    | 121
user1       | 09-08-2014    | 122
user1       | 09-09-2014    | 0
user1       | 09-10-2014    | 144
user1       | 09-11-2014    | 166
user2       | 09-01-2014    | 177
user2       | 09-04-2014    | 188
user2       | 09-05-2014    | 199
user2       | 09-06-2014    | 0
user2       | 09-07-2014    | 155

如果[UserCode]不是零,则应仅计算连续天数(如果为结果).
UserDate介于09-01-2014和09-11-2014之间.
仅在结果为2或更多时显示结果.

我希望我的sql查询返回的是：

UserName    | StartDate     | EndDate       | Result
----------------------------------------------------------
user1       | 09-01-2014    | 09-03-2014    | 2
user1       | 09-08-2014    | 09-11-2014    | 3
user2       | 09-04-2014    | 09-07-2014    | 3

这只能使用SQL查询吗？

解决方法

SELECT  UserName,UserDate,UserCode,GroupingSet = DATEADD(DAY,-ROW_NUMBER() OVER(PARTITION BY UserName 
                                                        ORDER BY UserDate),UserDate)
FROM    UserTable;

这给出了：

UserName    | UserDate      | UserCode   | GroupingSet
------------+---------------+------------+-------------
user1       | 09-01-2014    | 1          | 08-31-2014    
user1       | 09-02-2014    | 0          | 08-31-2014    
user1       | 09-03-2014    | 1          | 08-31-2014    
user1       | 09-08-2014    | 1          | 09-04-2014    
user1       | 09-09-2014    | 0          | 09-04-2014    
user1       | 09-10-2014    | 1          | 09-04-2014    
user1       | 09-11-2014    | 1          | 09-04-2014    
user2       | 09-01-2014    | 1          | 08-31-2014    
user2       | 09-04-2014    | 1          | 09-02-2014    
user2       | 09-05-2014    | 1          | 09-02-2014    
user2       | 09-06-2014    | 0          | 09-02-2014    
user2       | 09-07-2014    | 1          | 09-02-2014

如您所见,这为连续行的GroupingSet提供了一个常量值.然后,您可以按此列分组以获取所需的摘要：

WITH CTE AS
(   SELECT  UserName,-ROW_NUMBER() OVER(PARTITION BY UserName 
                                                            ORDER BY UserDate),UserDate)
    FROM    UserTable
)
SELECT  UserName,StartDate = MIN(UserDate),EndDate = MAX(UserDate),Result = COUNT(NULLIF(UserCode,0))
FROM    CTE
GROUP BY UserName,GroupingSet
HAVING COUNT(NULLIF(UserCode,0)) > 1
ORDER BY UserName,StartDate;

Example on SQL Fiddle

（编辑：开发网_新乡站长网）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!