Mysql 从sql中的事件日志表到累积快照事实的有效查询

Mysql 从sql中的事件日志表到累积快照事实的有效查询,mysql,sql-server,data-warehouse,Mysql,Sql Server,Data Warehouse,此示例是在SQLServer2016中构建的,但也应适用于MySQL 8.X 我将事件日志数据存储在带有以下示例数据的表fact\u user\u event\u activity中: event_date_key user_key step_key session_id event_timestamp 20140411 123 1 1000 2014-04-11 08:00:00.000 20140411

此示例是在SQLServer2016中构建的,但也应适用于MySQL 8.X

我将事件日志数据存储在带有以下示例数据的表
fact\u user\u event\u activity
中:

event_date_key  user_key    step_key    session_id  event_timestamp
20140411        123         1           1000        2014-04-11 08:00:00.000
20140411        123         2           1000        2014-04-11 08:10:00.000
20140411        123         3           1000        2014-04-11 08:20:00.000
20140411        123         4           1000        2014-04-11 08:30:00.000
20140411        125         1           1001        2014-04-11 09:10:00.000
20140411        123         5           1000        2014-04-11 08:31:00.000
20140411        125         2           1001        2014-04-11 09:30:00.000
20140411        125         3           1001        2014-04-11 09:50:00.000  <-- 
20140411        125         3           1001        2014-04-11 09:51:00.000  <--
20140411        125         4           1001        2014-04-11 09:52:00.000
这将用作累积快照的ETL查询

设置

DROP TABLE IF EXISTS  [fact_user_event_activity]
;
CREATE TABLE [fact_user_event_activity] (
  [event_date_key] INT DEFAULT NULL,
  [user_key] BIGINT NOT NULL,
  [step_key] BIGINT NOT NULL,
  [session_id] BIGINT NOT NULL,
  [event_timestamp] datetime NOT NULL
)
;
INSERT INTO [fact_user_event_activity]
VALUES (20140411, 123, 1, 1000, N'2014-04-11 08:00:00'),
(20140411, 123, 2, 1000, N'2014-04-11 08:10:00'),
(20140411, 123, 3, 1000, N'2014-04-11 08:20:00'),
(20140411, 123, 4, 1000, N'2014-04-11 08:30:00'),
(20140411, 125, 1, 1001, N'2014-04-11 09:10:00'),
(20140411, 123, 5, 1000, N'2014-04-11 08:31:00'),
(20140411, 125, 2, 1001, N'2014-04-11 09:30:00'),
(20140411, 125, 3, 1001, N'2014-04-11 09:50:00'),
(20140411, 125, 3, 1001, N'2014-04-11 09:51:00'),
(20140411, 125, 4, 1001, N'2014-04-11 09:52:00'),
(20140411, 129, 1, 1005, N'2014-04-11 09:08:00'),
(20140411, 129, 2, 1005, N'2014-04-11 09:10:00'),
(20140411, 129, 3, 1005, N'2014-04-11 09:12:00'),
(20140411, 129, 3, 1005, N'2014-04-11 09:13:00'),
(20140411, 129, 4, 1005, N'2014-04-11 09:14:00'),
(20140411, 129, 5, 1005, N'2014-04-11 09:18:00')
;
我的尝试

为了容易理解代码,我分两步来实现这一点:

  • 从开始(会话开始)获取每个步骤的持续时间
  • 计算每个步骤的持续时间从开始到结束之间的差值
  • 这将返回我所期望的结果,但我确信我可能会使事情变得过于复杂,并且会遇到约500万条记录,因此我想知道是否有更好的方法,或者我是否遗漏了一些东西

    -- Step 1
    -- to improve performance, use temp table instead of CTE
    -- Use TIMESTAMPDIFF in MySQL instead of DATEDIFF
    WITH durations_from_start_tmp AS
        (
        SELECT session_id, user_key, FIRST_VALUE(fuea.event_timestamp) OVER(PARTITION BY user_key, fuea.session_id ORDER BY fuea.event_timestamp) first_login,
        DENSE_RANK() OVER(PARTITION BY user_key, step_key, fuea.session_id ORDER BY fuea.event_timestamp) AS rnk,
        CASE WHEN step_key = 2 THEN DATEDIFF(MINUTE, FIRST_VALUE(fuea.event_timestamp) OVER(PARTITION BY user_key, fuea.session_id ORDER BY fuea.event_timestamp), fuea.event_timestamp) END AS step_1_duration_from_start,
        CASE WHEN step_key = 3 THEN DATEDIFF(MINUTE, FIRST_VALUE(fuea.event_timestamp) OVER(PARTITION BY user_key, fuea.session_id ORDER BY fuea.event_timestamp), fuea.event_timestamp) END AS step_2_duration_from_start,
        CASE WHEN step_key = 4 THEN DATEDIFF(MINUTE, FIRST_VALUE(fuea.event_timestamp) OVER(PARTITION BY user_key, fuea.session_id ORDER BY fuea.event_timestamp), fuea.event_timestamp) END AS step_3_duration_from_start,
        CASE WHEN step_key = 5 THEN DATEDIFF(MINUTE, FIRST_VALUE(fuea.event_timestamp) OVER(PARTITION BY user_key, fuea.session_id ORDER BY fuea.event_timestamp), fuea.event_timestamp) END AS step_4_duration_from_start
        FROM [fact_user_event_activity] fuea
        --WHERE event_timestamp > watermark --for incremental load
        )
    
    -- Step 2
    SELECT user_key, session_id, SUM(step_1_duration_from_start) AS step_1_duration_mins,
     SUM(step_2_duration_from_start) - SUM(step_1_duration_from_start) AS step_2_duration_mins ,
     SUM(step_3_duration_from_start) - SUM(step_2_duration_from_start) AS step_3_duration_mins ,
     SUM(step_4_duration_from_start) - SUM(step_3_duration_from_start) AS step_4_duration_mins
     FROM durations_from_start_tmp
     -- deals with repeated steps
     WHERE rnk = 1
     GROUP BY  user_key, session_id
    
    参考资料

    这可能与获得答案无关,但以防您不熟悉数据建模概念


    因此,您可能采取的一种方法是添加索引(假设您可以添加索引),例如:

    在[事实\用户\事件\活动](用户\键、会话\ id、步骤\键、事件\时间戳)上创建索引[SomeIndexName]

    (或者,如果您关心500m行上索引的大小,可以只执行一个include on step_键、event_timestamp。)

    然后跳过将窗口函数用于以下查询:

    SELECT user_key,
           session_id,
           step_1_duration = DATEDIFF(MINUTE, step_1_timestamp, step_2_timestamp),
           step_2_duration = DATEDIFF(MINUTE, step_2_timestamp, step_3_timestamp),
           step_3_duration = DATEDIFF(MINUTE, step_3_timestamp, step_4_timestamp),
           step_4_duration = DATEDIFF(MINUTE, step_4_timestamp, step_5_timestamp)
    FROM 
    (
        SELECT user_key, session_id,
               step_1_timestamp = MIN(CASE WHEN step_key = 1 THEN event_timestamp END),
               step_2_timestamp = MIN(CASE WHEN step_key = 2 THEN event_timestamp END),
               step_3_timestamp = MIN(CASE WHEN step_key = 3 THEN event_timestamp END),
               step_4_timestamp = MIN(CASE WHEN step_key = 4 THEN event_timestamp END),
               step_5_timestamp = MIN(CASE WHEN step_key = 5 THEN event_timestamp END)
        FROM fact_user_event_activity
        GROUP BY user_key, session_id
    ) AS T;
    

    (理论上只需进行索引扫描,无需进行任何排序。)

    我非常想说这不符合主题,更适合。。正如您所说的代码是有效的,stackoverflow更适用于有缺陷的代码。但是,在您需要使用窗口函数比较当前记录和“第一”记录时回答您的问题,很可能是性能最好的方法。@RaymondNijland非常感谢,我一直在到处寻找这类分析问题的例子,我认为这是“一个特定的编程问题”,因为“累积快照”似乎是一种非常普遍的做法,鉴于这是我第一次尝试使用这些事实,我担心复杂性,而且可能缺少一个未知的窗口函数。由于您在最终选择中似乎没有实际使用窗口函数,因此最好跳过使用它们。例如,您可以只选择每个步骤的最小值(例如,
    step\u 1\u timestamp=min(step\u key=1的情况下,事件\u timestamp结束),step\u 2\u timestamp=min(step\u key=2的情况下…
    ),然后通过执行类似于
    step\u 1\u duration=datediff(分钟,step\u 1\u timestamp,step\u 2\u timestamp)的datediff来获取每个步骤的持续时间
    。在您分组的列上有一个索引和步骤键/事件时间戳,这似乎比使用(不必要的)更好窗口功能。@ZLKthanks,我想我明白了,你能不能把这个方法作为一个答案,这样我就可以把你的整个想法形象化,当然,如果这恰好是答案的话,就投票表决,甚至把它标记为答案?这正是我想要的,更有意义。我的思维过程是一样的,但我不需要计算du从一开始定量供应=)这更简单。
    SELECT user_key,
           session_id,
           step_1_duration = DATEDIFF(MINUTE, step_1_timestamp, step_2_timestamp),
           step_2_duration = DATEDIFF(MINUTE, step_2_timestamp, step_3_timestamp),
           step_3_duration = DATEDIFF(MINUTE, step_3_timestamp, step_4_timestamp),
           step_4_duration = DATEDIFF(MINUTE, step_4_timestamp, step_5_timestamp)
    FROM 
    (
        SELECT user_key, session_id,
               step_1_timestamp = MIN(CASE WHEN step_key = 1 THEN event_timestamp END),
               step_2_timestamp = MIN(CASE WHEN step_key = 2 THEN event_timestamp END),
               step_3_timestamp = MIN(CASE WHEN step_key = 3 THEN event_timestamp END),
               step_4_timestamp = MIN(CASE WHEN step_key = 4 THEN event_timestamp END),
               step_5_timestamp = MIN(CASE WHEN step_key = 5 THEN event_timestamp END)
        FROM fact_user_event_activity
        GROUP BY user_key, session_id
    ) AS T;