Question

我有一个表，其中包含如下事件：ID，USER_ID，CREATED_AT，EVENT_NAME。

我正在尝试获取用户通常在会话中创建的事件序列。当用户事件与上一个事件相距5分钟以上时，新的会话开始。

我尽力能够创建一个具有以下信息的视图：

以该顺序读取表，每当“ TIME_DIFF”大于5分钟时，就会开始新的会话。

我现在该如何按会话汇总事件，以便最终获得类似的内容？

在表格下方，查看和一些测试数据：

CREATE SCHEMA test;

CREATE TABLE test."TRACKING_EVENTS" (
    "ID" serial PRIMARY key,
    "USER_ID" text,
    "CREATED_AT" TIMESTAMP,
    "EVENT_NAME" text
);

CREATE VIEW
    test."ORDERED_EVENTS"
AS
    SELECT 
        "ID", 
        "USER_ID", 
        "CREATED_AT" AS "EVENT_TIME", 
        "EVENT_NAME",
        CASE WHEN 
            lag("CREATED_AT", 1) OVER (ORDER BY "USER_ID", "CREATED_AT") < "CREATED_AT" 
        THEN 
            lag("CREATED_AT", 1) OVER (ORDER BY "USER_ID", "CREATED_AT")
        ELSE
            NULL 
        END AS "PREVIOUS_EVENT_TIME" 
    FROM 
        test."TRACKING_EVENTS";

CREATE VIEW
    test."ORDERED_EVENTS_WITH_DIFF"
AS
    SELECT  
        "ID", 
        "USER_ID", 
        "EVENT_TIME", 
        "EVENT_NAME",
        "PREVIOUS_EVENT_TIME",
        "EVENT_TIME" - "PREVIOUS_EVENT_TIME" AS "TIME_DIFF"
    FROM 
        test."ORDERED_EVENTS";

-- Period 1
INSERT INTO test."TRACKING_EVENTS" ("ID", "USER_ID", "CREATED_AT", "EVENT_NAME") 
VALUES (1, 'user1', '2019-1-1 01:00:00'::timestamp, 'EVENT_1');
INSERT INTO test."TRACKING_EVENTS" ("ID", "USER_ID", "CREATED_AT", "EVENT_NAME") 
VALUES (3, 'user1', '2019-1-1 01:00:05'::timestamp, 'EVENT_2');
INSERT INTO test."TRACKING_EVENTS" ("ID", "USER_ID", "CREATED_AT", "EVENT_NAME") 
VALUES (5, 'user1', '2019-1-1 01:00:10'::timestamp, 'EVENT_3');

INSERT INTO test."TRACKING_EVENTS" ("ID", "USER_ID", "CREATED_AT", "EVENT_NAME") 
VALUES (2, 'user2', '2019-1-1 01:00:01'::timestamp, 'EVENT_1');
INSERT INTO test."TRACKING_EVENTS" ("ID", "USER_ID", "CREATED_AT", "EVENT_NAME") 
VALUES (4, 'user2', '2019-1-1 01:00:06'::timestamp, 'EVENT_2');
INSERT INTO test."TRACKING_EVENTS" ("ID", "USER_ID", "CREATED_AT", "EVENT_NAME") 
VALUES (6, 'user2', '2019-1-1 01:00:11'::timestamp, 'EVENT_3');

-- Period 2
INSERT INTO test."TRACKING_EVENTS" ("ID", "USER_ID", "CREATED_AT", "EVENT_NAME") 
VALUES (7, 'user1', '2019-1-1 01:10:00'::timestamp, 'EVENT_1');
INSERT INTO test."TRACKING_EVENTS" ("ID", "USER_ID", "CREATED_AT", "EVENT_NAME") 
VALUES (9, 'user1', '2019-1-1 01:10:05'::timestamp, 'EVENT_2');
INSERT INTO test."TRACKING_EVENTS" ("ID", "USER_ID", "CREATED_AT", "EVENT_NAME") 
VALUES (11, 'user1', '2019-1-1 01:10:10'::timestamp, 'EVENT_3');

INSERT INTO test."TRACKING_EVENTS" ("ID", "USER_ID", "CREATED_AT", "EVENT_NAME") 
VALUES (8, 'user2', '2019-1-1 01:10:01'::timestamp, 'EVENT_1');
INSERT INTO test."TRACKING_EVENTS" ("ID", "USER_ID", "CREATED_AT", "EVENT_NAME") 
VALUES (10, 'user2', '2019-1-1 01:10:06'::timestamp, 'EVENT_2');
INSERT INTO test."TRACKING_EVENTS" ("ID", "USER_ID", "CREATED_AT", "EVENT_NAME") 
VALUES (12, 'user2', '2019-1-1 01:10:11'::timestamp, 'EVENT_3');

Answer 1

我想这就是你想要的：

select user_id, session,
       array_agg(event_name order by created_at)
from (select tt.*,
             count(*) filter (where prev_ca < created_at - interval '5 minute') over (partition by user_id order by created_at) as session
      from (select tt.*,
                   lag(created_at) over (partition by user_id order by CREATED_AT) as prev_ca
            from test."TRACKING_EVENTS" tt
           ) tt
     ) tt
group by user_id, session
order by user_id, session;

请注意，这使用array_agg()而不是string_agg()。您正在使用Postgres，因此array_agg()是将多个值组合在一起的好方法。

Postgres-按会话汇总用户事件

1 个答案: