我发现连续几天有很多stackoverflow QnAs 仍然答案太短，我无法理解发生了什么。
为了具体，我将组成一个模型（或表） （如果它有所作为，我正在使用postgresql。）
CREATE TABLE work (
id integer NOT NULL,
user_id integer NOT NULL,
arrived_at timestamp with time zone NOT NULL
);
insert into work(user_id, arrived_at) values(1, '01/03/2011');
insert into work(user_id, arrived_at) values(1, '01/04/2011');
（最简单的形式）对于给定的用户，我想找到最后连续的日期范围。
（我的最终目标）对于给定的用户，我想找到他连续的工作日 如果他昨天来上班，他仍然（截至今天）有连续工作的机会。所以我昨天连续几天给他看 但如果他昨天错过了，他的连续几天是0或1，取决于他今天是否来。
今天说是第8天。
3 * 5 6 7 * = 3 days (5 to 7)
3 * 5 6 7 8 = 4 days (5 to 8)
3 4 5 * 7 * = 1 day (7 to 7)
3 * * * * * = 0 day
3 * * * * 8 = 1 day (8 to 8)
答案 0 :(得分：3)
以下是使用CTE
WITH RECURSIVE CTE(attendanceDate)
AS
(
SELECT * FROM
(
SELECT attendanceDate FROM attendance WHERE attendanceDate = current_date
OR attendanceDate = current_date - INTERVAL '1 day'
ORDER BY attendanceDate DESC
LIMIT 1
) tab
UNION ALL
SELECT a.attendanceDate FROM attendance a
INNER JOIN CTE c
ON a.attendanceDate = c.attendanceDate - INTERVAL '1 day'
)
SELECT COUNT(*) FROM CTE;
以下是查询的工作方式：
attendance表中选择今天的记录。如果今天的记录不可用，那么它会选择昨天的记录如果您想选择最新的连续日期范围而不管用户最近的出勤时间（今天，昨天或前几天），那么CTE的初始化部分必须替换为以下代码段：
SELECT MAX(attendanceDate) FROM attendance
[编辑] 以下是SQL Fiddle的查询，它可以解决您的问题＃1：SQL Fiddle
答案 1 :(得分：0)
-- some data
CREATE table dayworked (
id SERIAL NOT NULL PRIMARY KEY
, user_id INTEGER NOT NULL
, arrived_at DATE NOT NULL
, UNIQUE (user_id, arrived_at)
);
INSERT INTO dayworked(user_id, arrived_at) VALUES
( 1, '2014-02-03')
,( 1, '2014-02-05')
,( 1, '2014-02-06')
,( 1, '2014-02-07')
--
,( 2, '2014-02-03')
,( 2, '2014-02-05')
,( 2, '2014-02-06')
,( 2, '2014-02-07')
,( 2, '2014-02-08')
--
,( 3, '2014-02-03')
,( 3, '2014-02-04')
,( 3, '2014-02-05')
,( 3, '2014-02-07')
--
,( 5, '2014-02-08')
;
-- The query
WITH RECURSIVE stretch AS (
SELECT dw.user_id AS user_id
, dw.arrived_at AS first_day
, dw.arrived_at AS last_day
, 1::INTEGER AS nday
FROM dayworked dw
WHERE NOT EXISTS ( -- Find start of chain: no previous day
SELECT * FROM dayworked nx
WHERE nx.user_id = dw.user_id
AND nx. arrived_at = dw.arrived_at -1
)
UNION ALL
SELECT dw.user_id AS user_id
, st.first_day AS first_day
, dw.arrived_at AS last_day
, 1+st.nday AS nday
FROM dayworked dw -- connect to chain: previous day := day before this day
JOIN stretch st ON st.user_id = dw.user_id AND st.last_day = dw.arrived_at -1
)
SELECT * FROM stretch st
WHERE (st.nday > 1 OR st.first_day = NOW()::date ) -- either more than one consecutive dat or starting today
AND NOT EXISTS ( -- Only the most recent stretch
SELECT * FROM stretch nx
WHERE nx.user_id = st .user_id
AND nx.first_day > st.first_day
)
AND NOT EXISTS ( -- omit partial chains
SELECT * FROM stretch nx
WHERE nx.user_id = st .user_id
AND nx.first_day = st.first_day
AND nx.last_day > st.last_day
)
;
结果：
CREATE TABLE
INSERT 0 14
user_id | first_day | last_day | nday
---------+------------+------------+------
1 | 2014-02-05 | 2014-02-07 | 3
2 | 2014-02-05 | 2014-02-08 | 4
(2 rows)
答案 2 :(得分：0)
您可以使用范围类型创建聚合：
Create function sfunc (tstzrange, timestamptz)
returns tstzrange
language sql strict as $$
select case when $2 - upper($1) <= '1 day'::interval
then tstzrange(lower($1), $2, '[]')
else tstzrange($2, $2, '[]') end
$$;
Create aggregate consecutive (timestamptz) (
sfunc = sfunc,
stype = tstzrange,
initcond = '[,]'
);
使用正确顺序的聚合获得最后到达的连续日期范围：
Select user_id, consecutive(arrived_at order by arrived_at)
from work
group by user_id;
┌─────────┬─────────────────────────────────────────────────────┐
│ user_id │ consecutive │
├─────────┼─────────────────────────────────────────────────────┤
│ 1 │ ["2011-01-03 00:00:00+02","2011-01-05 00:00:00+02"] │
│ 2 │ ["2011-01-06 00:00:00+02","2011-01-06 00:00:00+02"] │
└─────────┴─────────────────────────────────────────────────────┘
在窗口函数中使用聚合：
Select *,
consecutive(arrived_at)
over (partition by user_id order by arrived_at)
from work;
┌────┬─────────┬────────────────────────┬─────────────────────────────────────────────────────┐
│ id │ user_id │ arrived_at │ consecutive │
├────┼─────────┼────────────────────────┼─────────────────────────────────────────────────────┤
│ 1 │ 1 │ 2011-01-03 00:00:00+02 │ ["2011-01-03 00:00:00+02","2011-01-03 00:00:00+02"] │
│ 2 │ 1 │ 2011-01-04 00:00:00+02 │ ["2011-01-03 00:00:00+02","2011-01-04 00:00:00+02"] │
│ 3 │ 1 │ 2011-01-05 00:00:00+02 │ ["2011-01-03 00:00:00+02","2011-01-05 00:00:00+02"] │
│ 4 │ 2 │ 2011-01-06 00:00:00+02 │ ["2011-01-06 00:00:00+02","2011-01-06 00:00:00+02"] │
└────┴─────────┴────────────────────────┴─────────────────────────────────────────────────────┘
查询结果以找到所需内容：
With work_detail as (select *,
consecutive(arrived_at)
over (partition by user_id order by arrived_at)
from work)
select arrived_at, upper(consecutive) - lower(consecutive) as days
from work_detail
where user_id = 1 and upper(consecutive) != lower(consecutive)
order by arrived_at desc
limit 1;
┌────────────────────────┬────────┐
│ arrived_at │ days │
├────────────────────────┼────────┤
│ 2011-01-05 00:00:00+02 │ 2 days │
└────────────────────────┴────────┘
答案 3 :(得分：0)
你甚至可以在没有递归CTE的情况下做到这一点：
使用generate_series()，LEFT JOIN，row_count()和最终LIMIT 1：
1表示“今天”加上连续几天直到“昨天”：
SELECT count(*) -- 1 / 0 for "today"
+ COALESCE(( -- + optional count of consecutive days up until "yesterday"
SELECT ct
FROM (
SELECT d.ct, count(w.arrived_at) OVER (ORDER BY d.ct) AS day_ct
FROM generate_series(1, 8) AS d(ct) -- maximum = 8
LEFT JOIN work w ON w.arrived_at >= current_date - d.ct
AND w.arrived_at < current_date - (d.ct - 1)
AND w.user_id = 1 -- given user
) sub
WHERE ct = day_ct
ORDER BY ct DESC
LIMIT 1
), 0) AS total
FROM work
WHERE arrived_at >= current_date -- no future timestamps
AND user_id = 1 -- given user
假设每天0或1次进入。应该很快。
为获得最佳性能（对于此类或CTE解决方案），您将拥有多列索引，如：
CREATE INDEX foo_idx ON work (user_id,arrived_at);