按日期获取状态计数但只计算连续行

Question

我有这个数据：

ID  Name        Status  Date
1   Machine1    Active  2018-01-01
2   Machine2    Fault   2018-01-01
3   Machine3    Active  2018-01-01
4   Machine1    Fault   2018-01-02
5   Machine2    Active  2018-01-02
6   Machine3    Active  2018-01-02
7   Machine2    Active  2018-01-03
8   Machine1    Fault   2018-01-03
9   Machine2    Active  2018-01-04
10  Machine1    Fault   2018-01-04
11  Machine3    Active  2018-01-06

输入

我想要输出这些数据

预期输出

Name           Last Status  Count
Machine1         Fault       3  
Machine2         Active      3  
Machine3         Active      1       Because Date is not Continuous

*Count : 连续历史中的最后一个状态数

Answer 1

我认为这会起作用，尽管 SQLFiddle 目前身体不适，所以我无法测试：

SELECT [Name], [Status], ct as [Count]
FROM (
 SELECT 
  [name], 
  [status], 
  [date],
  1 + (SUM( grp ) OVER (PARTITION BY [name], [status] ORDER BY [date] ROWS BETWEEN 1 PRECEDING AND 0 FOLLOWING ) * grp) ct,
  row_number() over(partition by [name] order by [date] desc) rn
 FROM
 (
  SELECT *, CASE WHEN LAG([Date]) OVER(PARTITION BY [name], [status] ORDER BY [date] ) = DATEADD(day, -1, [date]) THEN 1 ELSE 0 END grp
  FROM t
 ) x
) y
WHERE
  rn = 1

它首先使用 LAG 查看当前行和前一行（将数据分组为机器名称和状态，按日期对数据排序），如果当前日期与前一个日期相差 1 天，则它记录 1 否则记录 0

这些 1 和 0 以运行ning 总的方式求和，当机器名称或状态改变时重置（sum() over() 的分区）

另外我们想只考虑机器名的数据，我们只想要每台机器的最新记录，所以我们按机器名分区，按日期降序计数，然后选择（使用 where 子句）每台机器编号为 1 的行

如果您运行单独查询，实际上更有意义，就像这样

计算 "is the current report consecutive with the previous report, for the given status and machine" 1 = 是，0 = 否：

SELECT *, CASE WHEN LAG([Date]) OVER(PARTITION BY [name], [status] ORDER BY [date] ) = DATEADD(day, -1, [date]) THEN 1 ELSE 0 END grp
  FROM t

计算 "what is the running total of the current block of consecutive reports":

SELECT 
  [name], 
  [status], 
  [date],
  1 + (SUM( grp ) OVER (PARTITION BY [name], [status] ORDER BY [date] ROWS BETWEEN 1 PRECEDING AND 0 FOLLOWING ) * grp) ct,
  row_number() over(partition by [name] order by [date] desc) rn
 FROM
 (
  SELECT *, CASE WHEN LAG([Date]) OVER(PARTITION BY [name], [status] ORDER BY [date] ) = DATEADD(day, -1, [date]) THEN 1 ELSE 0 END grp
  FROM t
 ) x

然后当然是整件事，但没有 where 子句，所以您可以看到我们正在丢弃的数据：

SELECT [Name], [Status], ct as [Count]
FROM (
 SELECT 
  [name], 
  [status], 
  [date],
  1 + (SUM( grp ) OVER (PARTITION BY [name], [status] ORDER BY [date] ROWS BETWEEN 1 PRECEDING AND 0 FOLLOWING ) * grp) ct,
  row_number() over(partition by [name] order by [date] desc) rn
 FROM
 (
  SELECT *, CASE WHEN LAG([Date]) OVER(PARTITION BY [name], [status] ORDER BY [date] ) = DATEADD(day, -1, [date]) THEN 1 ELSE 0 END grp
  FROM t
 ) x
) y

Fiddle终于醒了：

http://www.sqlfiddle.com/#!18/77dae/2

Answer 2

我相信就这么简单：

WITH cte1 AS (
    SELECT
        Name,
        Status,
        DATEADD(DAY, ROW_NUMBER() OVER (PARTITION BY Name, Status ORDER BY Date DESC) - 1, Date) AS GroupingDate
    FROM testdata
), cte2 AS (
    SELECT
        Name,
        Status,
        RANK() OVER (PARTITION BY Name ORDER BY GroupingDate DESC) AS GroupingNumber
    FROM cte1
)
SELECT Name, Status AS LastStatus, COUNT(*) AS LastStatusCount
FROM cte2
WHERE GroupingNumber = 1
GROUP BY Name, Status
ORDER BY Name

Result and DBFiddle:

| Name     | LastStatus | LastStatusCount |
|----------|------------|-----------------|
| Machine1 | Fault      | 3               |
| Machine2 | Active     | 3               |
| Machine3 | Active     | 1               |

为了了解这是如何工作的，请查看 CTE 生成的中间值：

| Name     | Status | Date                | RowNumber | GroupingDate        | GroupingNumber |
|----------|--------|---------------------|-----------|---------------------|----------------|
| Machine1 | Fault  | 04/01/2018 00:00:00 | 1         | 04/01/2018 00:00:00 | 1              |
| Machine1 | Fault  | 03/01/2018 00:00:00 | 2         | 04/01/2018 00:00:00 | 1              |
| Machine1 | Fault  | 02/01/2018 00:00:00 | 3         | 04/01/2018 00:00:00 | 1              |
| Machine1 | Active | 01/01/2018 00:00:00 | 1         | 01/01/2018 00:00:00 | 4              |
| Machine2 | Active | 04/01/2018 00:00:00 | 1         | 04/01/2018 00:00:00 | 1              |
| Machine2 | Active | 03/01/2018 00:00:00 | 2         | 04/01/2018 00:00:00 | 1              |
| Machine2 | Active | 02/01/2018 00:00:00 | 3         | 04/01/2018 00:00:00 | 1              |
| Machine2 | Fault  | 01/01/2018 00:00:00 | 1         | 01/01/2018 00:00:00 | 4              |
| Machine3 | Active | 06/01/2018 00:00:00 | 1         | 06/01/2018 00:00:00 | 1              |
| Machine3 | Active | 02/01/2018 00:00:00 | 2         | 03/01/2018 00:00:00 | 2              |
| Machine3 | Active | 01/01/2018 00:00:00 | 3         | 03/01/2018 00:00:00 | 2              |

这里的诀窍是，如果两个数字是连续的，那么从它们中减去连续的数字将得到相同的值。例如。如果我们有 5, 6, 8, 9 那么按这个顺序减去 1, 2, 3, 4 将产生 4, 4, 5, 5.

按日期获取状态计数但只计算连续行

Get the count of statuses by date but only count continuous rows

sql-server

date

sql-server-2016

continuous