Question

PersistentId     UserId      EnterDate
111              1           June 1, 2015 17:05
112              1           June 1, 2015 17:21
113              1           June 1, 2015 17:27
114              1           June 1, 2015 18:25
115              1           June 1, 2015 19:00
116              2           June 1, 2015 18:05
117              2           June 1, 2015 18:21
118              2           June 1, 2015 19:27

我想得到一个UserId列表和每个UserId的计数，这样只有EnterDates＆lt;之间存在差异的行。包括30分钟。

因此对于上述数据，输出将是

UserId        Count
1             3
2             2

应为UserId 1提取的行包含persistentIds 111,114,115。

应为UserId 2提取的行包含persistentIds 116,118

关于如何编写此SQL查询的任何想法？

Answer 1

您的问题措辞不明确，但根据您的预期结果，我认为您希望使用NOT EXISTS过滤掉具有相同用户ID的另一条记录后不到30分钟的记录。像这样：

with d as ( 
SELECT 111 persistent_id, 1 user_id,  to_date('June 1, 2015 17:05','Month DD, YYYY HH24:MI') enter_date from dual UNION ALL
SELECT 112 persistent_id, 1 user_id,  to_date('June 1, 2015 17:21','Month DD, YYYY HH24:MI') from dual UNION ALL
SELECT 113 persistent_id, 1 user_id,  to_date('June 1, 2015 17:27','Month DD, YYYY HH24:MI') from dual UNION ALL
SELECT 114 persistent_id, 1 user_id,  to_date('June 1, 2015 18:25','Month DD, YYYY HH24:MI') from dual UNION ALL
SELECT 115 persistent_id, 1 user_id,  to_date('June 1, 2015 19:00','Month DD, YYYY HH24:MI') from dual UNION ALL
SELECT 116 persistent_id, 2 user_id,  to_date('June 1, 2015 18:05','Month DD, YYYY HH24:MI') from dual UNION ALL
SELECT 117 persistent_id, 2 user_id,  to_date('June 1, 2015 18:21','Month DD, YYYY HH24:MI') from dual UNION ALL
SELECT 118 persistent_id, 2 user_id,  to_date('June 1, 2015 19:27','Month DD, YYYY HH24:MI') from dual
)
select d.user_id, count(*)
from d
where not exists ( SELECT 'record for same userid but less than 30 minutes earlier'
                   FROM   d d2
                   WHERE  d2.user_id = d.user_id
                   AND    d2.enter_date between d.enter_date - (0.5/24) and d.enter_date 
                   and    d2.persistent_id != d.persistent_id )
group by d.user_id                   
order by d.user_id

Answer 2

您可以使用LAG功能获取上一个功能。为用户输入enterDate并计算事件之间的持续时间

select user_id, count(*)
from
(with d as ( 
SELECT 111 persistent_id, 1 user_id,  to_date('June 1, 2015 17:05','Month DD, YYYY HH24:MI') enter_date from dual UNION ALL
SELECT 112 persistent_id, 1 user_id,  to_date('June 1, 2015 17:21','Month DD, YYYY HH24:MI') from dual UNION ALL
SELECT 113 persistent_id, 1 user_id,  to_date('June 1, 2015 17:27','Month DD, YYYY HH24:MI') from dual UNION ALL
SELECT 114 persistent_id, 1 user_id,  to_date('June 1, 2015 18:25','Month DD, YYYY HH24:MI') from dual UNION ALL
SELECT 115 persistent_id, 1 user_id,  to_date('June 1, 2015 19:00','Month DD, YYYY HH24:MI') from dual UNION ALL
SELECT 116 persistent_id, 2 user_id,  to_date('June 1, 2015 18:05','Month DD, YYYY HH24:MI') from dual UNION ALL
SELECT 117 persistent_id, 2 user_id,  to_date('June 1, 2015 18:21','Month DD, YYYY HH24:MI') from dual UNION ALL
SELECT 118 persistent_id, 2 user_id,  to_date('June 1, 2015 19:27','Month DD, YYYY HH24:MI') from dual
)
select d.user_id, persistent_id, enter_date
,lag(persistent_id) over (partition by user_id  order by enter_date)
,lag(enter_date) over (partition by user_id  order by enter_date)
,(enter_date - nvl (lag(enter_date) over (partition by user_id  order by enter_date), enter_date))*24*60 duration
from d
) where duration < 30
group by user_id

--results
    USER_ID COUNT(*)
1   1   3
2   2   2

Answer 3

两个查询既提供了您的预期结果，又使用了30分钟的窗口但对您的要求有完全不同的解释......您可能想澄清这个问题。

SQL Fiddle

Oracle 11g R2架构设置：

CREATE TABLE table_name (PersistentId, UserId, EnterDate ) AS
          SELECT 111, 1,  to_date('June 1, 2015 17:05','Month DD, YYYY HH24:MI') FROM DUAL
UNION ALL SELECT 112, 1,  to_date('June 1, 2015 17:21','Month DD, YYYY HH24:MI') FROM DUAL
UNION ALL SELECT 113, 1,  to_date('June 1, 2015 17:27','Month DD, YYYY HH24:MI') FROM DUAL
UNION ALL SELECT 114, 1,  to_date('June 1, 2015 18:25','Month DD, YYYY HH24:MI') FROM DUAL
UNION ALL SELECT 115, 1,  to_date('June 1, 2015 19:00','Month DD, YYYY HH24:MI') FROM DUAL
UNION ALL SELECT 116, 2,  to_date('June 1, 2015 18:05','Month DD, YYYY HH24:MI') FROM DUAL
UNION ALL SELECT 117, 2,  to_date('June 1, 2015 18:21','Month DD, YYYY HH24:MI') FROM DUAL
UNION ALL SELECT 118, 2,  to_date('June 1, 2015 19:27','Month DD, YYYY HH24:MI') FROM DUAL

查询1 - 计算30分钟窗口的结果：

SELECT UserId,
       "Count"
FROM (
  SELECT UserID,
         COUNT(*) OVER ( PARTITION BY UserId ORDER BY EnterDate RANGE BETWEEN INTERVAL '30' MINUTE PRECEDING AND CURRENT ROW ) AS "Count",
         EnterDate,
         LEAD(EnterDate) OVER ( PARTITION BY UserId ORDER BY EnterDate ) AS nextEnterDate
  FROM   Table_Name
)
WHERE "Count" > 1
AND   EnterDate + INTERVAL '30' MINUTE < nextEnterDate

<强> Results ：

| USERID | Count |
|--------|-------|
|      1 |     3 |
|      2 |     2 |

查询2 - 计算另一行30分钟内的所有行：

SELECT UserID,
       COUNT(1) AS "Count"
FROM (
  SELECT UserID,
         EnterDate,
         LAG(EnterDate) OVER ( PARTITION BY UserId ORDER BY EnterDate ) AS prevDate,
         LEAD(EnterDate) OVER ( PARTITION BY UserId ORDER BY EnterDate ) AS nextDate
  FROM   Table_Name
)
WHERE  EnterDate - INTERVAL '30' MINUTE < prevDate
OR     EnterDate + INTERVAL '30' MINUTE > nextDate
GROUP BY UserId

<强> Results ：

| USERID | Count |
|--------|-------|
|      1 |     3 |
|      2 |     2 |

sql count其中diff Dates少于30分钟的行

3 个答案: