SQL 计数连续天数

Question

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，则只应计算连续天数（作为结果）。 UserDate 介于 09-01-2014 和 09-11-2014 之间。 仅当 Result 为 2 或更多时才显示结果。

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

这是否可以仅使用 SQL 查询？

Answer 1

这是一个间隙和岛屿问题。 解决这个问题的最简单方法是使用ROW_NUMBER()来识别序列中的间隙：

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

这给出：

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，这在GroupingSet为连续行提供了一个常量值。 然后，您可以按此列分组以获得所需的摘要：

WITH CTE AS
(   SELECT  UserName,
            UserDate,
            UserCode,
            GroupingSet = DATEADD(DAY, 
                                -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;

SQL Fiddle 示例

Answer 2

请尝试：

;with T1 as(
    select 
        *, 
        ROW_NUMBER() over (  order by UserName, UserDate) ID 
    from tbl
)
,T as (
    SELECT *, 1 CNT FROM T1 where ID=1
    union all
    SELECT b.*, (case when T.UserDate+1=b.UserDate and 
                           T.UserName=b.UserName then t.CNT 
                        else T.CNT+1 end)
    from T1 b INNER JOIN T on b.ID=T.ID+1
)
select distinct UserName, MIN(UserDate), max(UserDate)
,sum(case UserCode when 0 then 0 else 1 end) From T group by UserName, CNT
having COUNT(*)>1

SQL 小提琴演示