[英]SQL/Teradata: Return records where value in consecutive rows is the same
[英]SQL return consecutive records
一個簡單的表:
ForumPost
--------------
ID (int PK)
UserID (int FK)
Date (datetime)
我要返回的是特定用戶連續n天每天至少發表1次帖子的次數。
例:
User 15844 has posted at least 1 post a day for 30 consecutive days 10 times
我已經用linq / lambda標記了這個問題,還有一個很好的解決方案。 我知道我可以通過迭代所有用戶記錄來解決此問題,但這很慢。
您可以使用ROW_NUMBER()
查找一個連續的條目,想像一下下面的日期集及其row_number(從0開始):
Date RowNumber
20130401 0
20130402 1
20130403 2
20130404 3
20130406 4
20130407 5
對於連續的條目,如果從值中減去row_number,則會得到相同的結果。 例如
Date RowNumber date - row_number
20130401 0 20130401
20130402 1 20130401
20130403 2 20130401
20130404 3 20130401
20130406 4 20130402
20130407 5 20130402
然后,您可以按date - row_number
進行分組,以獲取連續的日期集(即前4條記錄和后2條記錄)。
要將其應用於您的示例,請使用:
WITH Posts AS
( SELECT FirstPost = DATEADD(DAY, 1 - ROW_NUMBER() OVER(PARTITION BY UserID ORDER BY [Date]), [Date]),
UserID,
Date
FROM ( SELECT DISTINCT UserID, [Date] = CAST(Date AS [Date])
FROM ForumPost
) fp
), Posts2 AS
( SELECT FirstPost,
UserID,
Days = COUNT(*),
LastDate = MAX(Date)
FROM Posts
GROUP BY FirstPost, UserID
)
SELECT UserID, ConsecutiveDates = MAX(Days)
FROM Posts2
GROUP BY UserID;
SQL Fiddle上的示例(簡單,每個用戶最多只能連續幾天)
編輯
我認為以上內容並不能完全回答問題,這將給出用戶發布的次數,或連續n天以上的發布次數:
WITH Posts AS
( SELECT FirstPost = DATEADD(DAY, 1 - ROW_NUMBER() OVER(PARTITION BY UserID ORDER BY [Date]), [Date]),
UserID,
Date
FROM ( SELECT DISTINCT UserID, [Date] = CAST(Date AS [Date])
FROM ForumPost
) fp
), Posts2 AS
( SELECT FirstPost,
UserID,
Days = COUNT(*),
FirstDate = MIN(Date),
LastDate = MAX(Date)
FROM Posts
GROUP BY FirstPost, UserID
)
SELECT UserID, [Times Over N Days] = COUNT(*)
FROM Posts2
WHERE Days >= 30
GROUP BY UserID;
我認為您的特定應用程序使這一過程變得非常簡單。 如果您在“ n”天的間隔中有“ n”個不同的日期,則這些“ n”個不同的日期必須是連續的。
滾動到底部,以獲取僅需要公用表表達式並更改為PostgreSQL的常規解決方案。 (開玩笑。由於時間緊迫,我在PostgreSQL中實現了。)
create table ForumPost (
ID integer primary key,
UserID integer not null,
post_date date not null
);
insert into forumpost values
(1, 1, '2013-01-15'),
(2, 1, '2013-01-16'),
(3, 1, '2013-01-17'),
(4, 1, '2013-01-18'),
(5, 1, '2013-01-19'),
(6, 1, '2013-01-20'),
(7, 1, '2013-01-21'),
(11, 2, '2013-01-15'),
(12, 2, '2013-01-16'),
(13, 2, '2013-01-17'),
(16, 2, '2013-01-17'),
(14, 2, '2013-01-18'),
(15, 2, '2013-01-19'),
(21, 3, '2013-01-17'),
(22, 3, '2013-01-17'),
(23, 3, '2013-01-17'),
(24, 3, '2013-01-17'),
(25, 3, '2013-01-17'),
(26, 3, '2013-01-17'),
(27, 3, '2013-01-17');
現在,讓我們看一下該查詢的輸出。 為簡便起見,我正在查看5天間隔,而不是30天間隔。
select userid, count(distinct post_date) distinct_dates
from forumpost
where post_date between '2013-01-15' and '2013-01-19'
group by userid;
USERID DISTINCT_DATES
1 5
2 5
3 1
對於符合條件的用戶,該5天間隔內的不同日期數必須為5,對嗎? 因此,我們只需要將該邏輯添加到HAVING子句中即可。
select userid, count(distinct post_date) distinct_dates
from forumpost
where post_date between '2013-01-15' and '2013-01-19'
group by userid
having count(distinct post_date) = 5;
USERID DISTINCT_DATES
1 5
2 5
更一般的解決方案
這么說真的沒有任何意義,如果您每天從2013-01-01到2013-01-31發布,那么您已經連續30天發布了2次。 相反,我希望時鍾從2013年1月31日開始。 我很抱歉在PostgreSQL中實現; 稍后我將嘗試在T-SQL中實現。
with first_posts as (
select userid, min(post_date) first_post_date
from forumpost
group by userid
),
period_intervals as (
select userid, first_post_date period_start,
(first_post_date + interval '4' day)::date period_end
from first_posts
), user_specific_intervals as (
select
userid,
(period_start + (n || ' days')::interval)::date as period_start,
(period_end + (n || ' days')::interval)::date as period_end
from period_intervals, generate_series(0, 30, 5) n
)
select userid, period_start, period_end,
(select count(distinct post_date)
from forumpost
where forumpost.post_date between period_start and period_end
and userid = forumpost.userid) distinct_dates
from user_specific_intervals
order by userid, period_start;
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