如何按日期查找一行中连续出现的数据？

问题描述 投票：0回答：1

我试图解决的问题是我需要知道 EmployeeId + TenantId (1A) 在连续运行中出现了多少次。我真的只关心它是否大于 1。

我的应用程序在工作日运行并插入行。它记录当前运行日期/时间，并且还添加上次运行的日期和时间（显然是从不同的表检索的）。

桌子

CREATE TABLE EmployeeHistory
(
   ID INTEGER PRIMARY KEY AUTOINCREMENT,
   CurrentRun TEXT,
   LastRun TEXT,
   EmployeeId TEXT,
   TenantId TEXT
);

样本数据

DELETE FROM EmployeeHistory;

INSERT INTO EmployeeHistory ('CurrentRun', 'LastRun', 'EmployeeId', 'TenantId')

VALUES

-- August 19th run
( '2023-08-19 00:00:00.000000', '2023-08-18 00:00:00.000000', '1', 'A' ), -- Consecutive! (Employee exists in August 18th run)
( '2023-08-19 00:00:00.000000', '2023-08-18 00:00:00.000000', '3', 'A' ), -- Should not be included because was not in August 18th run
( '2023-08-19 00:00:00.000000', '2023-08-18 00:00:00.000000', '1', 'B' ), -- Should not be included because was not in August 18th run
( '2023-08-19 00:00:00.000000', '2023-08-18 00:00:00.000000', '2', 'B' ), -- Consecutive! (Employee exists in August 18th run)
( '2023-08-19 00:00:00.000000', '2023-08-18 00:00:00.000000', '4', 'A' ), -- Consecutive! (Employee exists in all runs)
( '2023-08-19 00:00:00.000000', '2023-08-18 00:00:00.000000', '2', 'A' ), -- Consecutive! (Employee exists in August 18th run)

-- August 18th run
( '2023-08-18 00:00:00.000000', '2023-08-17 00:00:00.000000', '1', 'A' ), -- Consecutive (Employee exists in August 19th run)!
( '2023-08-18 00:00:00.000000', '2023-08-17 00:00:00.000000', '2', 'A' ), -- Consecutive (Employee exists in August 19th run)!
( '2023-08-18 00:00:00.000000', '2023-08-17 00:00:00.000000', '2', 'B' ), -- Consecutive (Employee exists in August 19th run)!
( '2023-08-18 00:00:00.000000', '2023-08-17 00:00:00.000000', '4', 'A' ), -- Consecutive! (Employee exists in all runs)

-- August 17th run
( '2023-08-17 00:00:00.000000', '2023-08-17 00:00:00.000000', '3', 'A' ), -- Should not be included because was not in August 18th run
( '2023-08-17 00:00:00.000000', '2023-08-17 00:00:00.000000', '5', 'A' ), -- Should not be included because was not in August 18th run
( '2023-08-17 00:00:00.000000', '2023-08-17 00:00:00.000000', '6', 'A' ), -- Should not be included because was not in August 18th run
( '2023-08-18 00:00:00.000000', '2023-08-17 00:00:00.000000', '4', 'A' ); -- Consecutive! (Employee exists in all runs)

电流输出：

ID  CurrentRun                  LastRun                     EmployeeId  TenantId
1   2023-08-19 00:00:00.000000  2023-08-18 00:00:00.000000  1           A
2   2023-08-19 00:00:00.000000  2023-08-18 00:00:00.000000  3           A
3   2023-08-19 00:00:00.000000  2023-08-18 00:00:00.000000  1           B
4   2023-08-19 00:00:00.000000  2023-08-18 00:00:00.000000  2           B
5   2023-08-19 00:00:00.000000  2023-08-18 00:00:00.000000  4           A
6   2023-08-19 00:00:00.000000  2023-08-18 00:00:00.000000  2           A
7   2023-08-18 00:00:00.000000  2023-08-17 00:00:00.000000  1           A
8   2023-08-18 00:00:00.000000  2023-08-17 00:00:00.000000  2           A
9   2023-08-18 00:00:00.000000  2023-08-17 00:00:00.000000  2           B
10  2023-08-18 00:00:00.000000  2023-08-17 00:00:00.000000  4           A
11  2023-08-17 00:00:00.000000  2023-08-17 00:00:00.000000  3           A
12  2023-08-17 00:00:00.000000  2023-08-17 00:00:00.000000  5           A
13  2023-08-17 00:00:00.000000  2023-08-17 00:00:00.000000  6           A
14  2023-08-18 00:00:00.000000  2023-08-17 00:00:00.000000  4           A

我用谷歌搜索并阅读了一些有关 LEAD 和 LAG 的内容，但我不知道如何取回我想要的数据。

我想到的一个解决方案是添加一列来跟踪连续发生的情况，并根据上次运行简单地增加给定员工的总和。这是一个不好的方法吗？

sql sqlite tsql gaps-and-islands
1个回答
0
投票
CREATE TABLE EmployeeHistory
    (
       ID INTEGER PRIMARY KEY AUTOINCREMENT,
       CurrentRun TEXT,
       LastRun TEXT,
       EmployeeId TEXT,
       TenantId TEXT
    );




    INSERT INTO EmployeeHistory ('CurrentRun', 'LastRun', 'EmployeeId', 'TenantId')
    
    VALUES
    
    -- August 19th run
    ( '2023-08-19 00:00:00.000000', '2023-08-18 00:00:00.000000', '1', 'A' ), -- Consecutive! (Employee exists in August 18th run)
    ( '2023-08-19 00:00:00.000000', '2023-08-18 00:00:00.000000', '3', 'A' ), -- Should not be included because was not in August 18th run
    ( '2023-08-19 00:00:00.000000', '2023-08-18 00:00:00.000000', '1', 'B' ), -- Should not be included because was not in August 18th run
    ( '2023-08-19 00:00:00.000000', '2023-08-18 00:00:00.000000', '2', 'B' ), -- Consecutive! (Employee exists in August 18th run)
    ( '2023-08-19 00:00:00.000000', '2023-08-18 00:00:00.000000', '4', 'A' ), -- Consecutive! (Employee exists in all runs)
    ( '2023-08-19 00:00:00.000000', '2023-08-18 00:00:00.000000', '2', 'A' ), -- Consecutive! (Employee exists in August 18th run)
    
    -- August 18th run
    ( '2023-08-18 00:00:00.000000', '2023-08-17 00:00:00.000000', '1', 'A' ), -- Consecutive (Employee exists in August 19th run)!
    ( '2023-08-18 00:00:00.000000', '2023-08-17 00:00:00.000000', '2', 'A' ), -- Consecutive (Employee exists in August 19th run)!
    ( '2023-08-18 00:00:00.000000', '2023-08-17 00:00:00.000000', '2', 'B' ), -- Consecutive (Employee exists in August 19th run)!
    ( '2023-08-18 00:00:00.000000', '2023-08-17 00:00:00.000000', '4', 'A' ), -- Consecutive! (Employee exists in all runs)
    
    -- August 17th run
    ( '2023-08-17 00:00:00.000000', '2023-08-17 00:00:00.000000', '3', 'A' ), -- Should not be included because was not in August 18th run
    ( '2023-08-17 00:00:00.000000', '2023-08-17 00:00:00.000000', '5', 'A' ), -- Should not be included because was not in August 18th run
    ( '2023-08-17 00:00:00.000000', '2023-08-17 00:00:00.000000', '6', 'A' ), -- Should not be included because was not in August 18th run
    ( '2023-08-18 00:00:00.000000', '2023-08-17 00:00:00.000000', '4', 'A' ); -- Consecutive! (Employee exists in all runs);




WITH ConsecutiveRuns AS (
  SELECT
    a.CurrentRun AS CurrentRunDate,
    b.CurrentRun AS LastRunDate,
    a.EmployeeId,
    a.TenantId
  FROM
    EmployeeHistory a
    JOIN EmployeeHistory b
  ON
    a.LastRun = b.CurrentRun AND
    a.EmployeeId = b.EmployeeId AND
    a.TenantId = b.TenantId
)

SELECT 
  EmployeeId,
  TenantId,
  COUNT(*) AS ConsecutiveCount
FROM 
  ConsecutiveRuns
GROUP BY 
  EmployeeId,
  TenantId
HAVING 
  ConsecutiveCount > 1;

|员工 ID |租户 ID |连续计数 | | :------------|:---------|:-----------------| | 4 |一个 | 2 | 小提琴

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