SQL问题 - 计算最大天数序列

Question

有一个访问数据表：

uid (INT) | created_at (DATETIME)

我想查找用户连续多少天访问过我们的应用。 例如：

SELECT DISTINCT DATE(created_at) AS d FROM visits WHERE uid = 123

将返回：

     d      
------------
 2012-04-28
 2012-04-29
 2012-04-30
 2012-05-03
 2012-05-04

有5个记录和两个间隔 - 3天（4月28日至30日）和2天（5月3日至4日）。

我的问题是如何找到用户连续访问应用程序的最大天数（示例中为3天）。 试图在SQL文档中找到合适的函数，但没有成功。 我错过了什么吗？

---

UPD：谢谢你们的回答！ 实际上，我正在使用vertica分析数据库（http://vertica.com/），但这是一个非常罕见的解决方案，只有少数人有使用它的经验。 虽然它支持SQL-99标准。

那么，大多数解决方案只需稍作修改即可。 最后我创建了自己的查询版本：

-- returns starts of the vitit series 
SELECT t1.d as s FROM testing t1
LEFT JOIN testing t2 ON DATE(t2.d) = DATE(TIMESTAMPADD('day', -1, t1.d))
WHERE t2.d is null GROUP BY t1.d

          s          
---------------------
 2012-04-28 01:00:00
 2012-05-03 01:00:00

-- returns end of the vitit series 
SELECT t1.d as f FROM testing t1
LEFT JOIN testing t2 ON DATE(t2.d) = DATE(TIMESTAMPADD('day', 1, t1.d))
WHERE t2.d is null GROUP BY t1.d

          f          
---------------------
 2012-04-30 01:00:00
 2012-05-04 01:00:00

所以现在我们只需要以某种方式加入它们，例如通过行索引。

SELECT s, f, DATEDIFF(day, s, f) + 1 as seq FROM (
    SELECT t1.d as s, ROW_NUMBER() OVER () as o1 FROM testing t1
    LEFT JOIN testing t2 ON DATE(t2.d) = DATE(TIMESTAMPADD('day', -1, t1.d))
    WHERE t2.d is null GROUP BY t1.d
) tbl1 LEFT JOIN (
    SELECT t1.d as f, ROW_NUMBER() OVER () as o2 FROM testing t1
    LEFT JOIN testing t2 ON DATE(t2.d) = DATE(TIMESTAMPADD('day', 1, t1.d))
    WHERE t2.d is null GROUP BY t1.d
) tbl2 ON o1 = o2 

样本输出：

          s          |          f          | seq 
---------------------+---------------------+-----
 2012-04-28 01:00:00 | 2012-04-30 01:00:00 |   3
 2012-05-03 01:00:00 | 2012-05-04 01:00:00 |   2

Answer 1

另一种方法，最短，做自我加入：

with grouped_result as
(
    select 
       sr.d,
       sum((fr.d is null)::int) over(order by sr.d) as group_number
    from tbl sr
    left join tbl fr on sr.d = fr.d + interval '1 day'
)
select d, group_number, count(d) over m as consecutive_days
from grouped_result
window m as (partition by group_number)

输出：

          d          | group_number | consecutive_days 
---------------------+--------------+------------------
 2012-04-28 08:00:00 |            1 |                3
 2012-04-29 08:00:00 |            1 |                3
 2012-04-30 08:00:00 |            1 |                3
 2012-05-03 08:00:00 |            2 |                2
 2012-05-04 08:00:00 |            2 |                2
(5 rows)

现场测试： http ： //www.sqlfiddle.com/#！1/93789/1

sr =第二行，fr =第一行（或者可能是前一行？ ツ ）。 基本上我们正在进行反向跟踪，这是一个不支持LAG数据库的模拟延迟（Postgres支持LAG，但解决方案很长 ，因为窗口不支持嵌套窗口）。 所以在这个查询中，我们使用混合方法，通过join模拟LAG，然后对它使用SUM窗口，这会产生组号

UPDATE

忘了把最后的查询，上面的查询说明了组编号的基础，需要将其变形为：

with grouped_result as
(
    select 
       sr.d,
       sum((fr.d is null)::int) over(order by sr.d) as group_number
    from tbl sr
    left join tbl fr on sr.d = fr.d + interval '1 day'
)
select min(d) as starting_date, max(d) as end_date, count(d) as consecutive_days
from grouped_result
group by group_number
-- order by consecutive_days desc limit 1


STARTING_DATE                END_DATE                     CONSECUTIVE_DAYS
April, 28 2012 08:00:00-0700 April, 30 2012 08:00:00-0700 3
May, 03 2012 08:00:00-0700   May, 04 2012 08:00:00-0700   2

UPDATE

我知道为什么我使用窗口函数的其他解决方案变得很长，我试图说明组编号的逻辑和计数组的时间变得很长。 如果我像我的MySql方法一样切入追逐，那么窗口函数可能会更短。 话虽如此，这是我的旧窗口功能方法，虽然现在更好：

with headers as
(
    select 
      d,lag(d) over m is null or d - lag(d) over m  <> interval '1 day' as header
    from tbl
    window m as (order by d)
)      
,sequence_group as
(
    select d, sum(header::int) over (order by d) as group_number
    from headers  
)
select min(d) as starting_date,max(d) as ending_date,count(d) as consecutive_days
from sequence_group
group by group_number
-- order by consecutive_days desc limit 1

现场测试： http ： //www.sqlfiddle.com/#！1/93789/21

Answer 2

在MySQL中你可以这样做：

SET @nextDate = CURRENT_DATE;
SET @RowNum = 1;

SELECT MAX(RowNumber) AS ConecutiveVisits
FROM    (   SELECT  @RowNum := IF(@NextDate = Created_At, @RowNum + 1, 1) AS RowNumber,
                    Created_At,
                    @NextDate := DATE_ADD(Created_At, INTERVAL 1 DAY) AS NextDate
            FROM    Visits
            ORDER BY Created_At
        ) Visits

这里的例子：

http://sqlfiddle.com/#!2/6e035/8

但是我并不是100％肯定这是最好的方法。

在Postgresql中：

 ;WITH RECURSIVE VisitsCTE AS
 (  SELECT  Created_At, 1 AS ConsecutiveDays
    FROM    Visits
    UNION ALL
    SELECT  v.Created_At, ConsecutiveDays + 1
    FROM    Visits v
            INNER JOIN VisitsCTE cte
                ON 1 + cte.Created_At = v.Created_At
)
SELECT  MAX(ConsecutiveDays) AS ConsecutiveDays
FROM    VisitsCTE

这里的例子：

http://sqlfiddle.com/#!1/16c90/9

Answer 3

我知道Postgresql有类似于MSSQL中可用的常用表表达式。 我对Postgresql并不熟悉，但下面的代码适用于MSSQL，可以满足您的需求。

create table #tempdates (
    mydate date
)

insert into #tempdates(mydate) values('2012-04-28')
insert into #tempdates(mydate) values('2012-04-29')
insert into #tempdates(mydate) values('2012-04-30')
insert into #tempdates(mydate) values('2012-05-03')
insert into #tempdates(mydate) values('2012-05-04');

with maxdays (s, e, c)
as
(
    select mydate, mydate, 1
    from #tempdates
    union all
    select m.s, mydate, m.c + 1
    from #tempdates t
    inner join maxdays m on DATEADD(day, -1, t.mydate)=m.e
)
select MIN(o.s),o.e,max(o.c)
from (
  select m1.s,max(m1.e) e,max(m1.c) c
  from maxdays m1
  group by m1.s
) o
group by o.e

drop table #tempdates

这里是SQL小提琴： http ：//sqlfiddle.com/#！3/ 42b38/2

Answer 4

所有这些都是非常好的答案，但我认为我应该通过展示另一种利用Vertica特有的分析能力的方法做出贡献（毕竟它是你付出的一部分）。 我保证最后的查询很简短。

首先，使用conditional_true_event（）进行查询。 从Vertica的文档：

为每行指定一个事件窗口编号，从0开始，当布尔参数表达式的结果为真时，将该数字递增1。

示例查询如下所示：

select uid, created_at, 
       conditional_true_event( created_at - lag(created_at) > '1 day' ) 
       over (partition by uid order by created_at) as seq_id
from visits;

并输出：

uid  created_at           seq_id  
---  -------------------  ------  
123  2012-04-28 00:00:00  0       
123  2012-04-29 00:00:00  0       
123  2012-04-30 00:00:00  0       
123  2012-05-03 00:00:00  1       
123  2012-05-04 00:00:00  1       
123  2012-06-04 00:00:00  2       
123  2012-06-04 00:00:00  2     

现在最后的查询变得简单了：

select uid, seq_id, count(1) num_days, min(created_at) s, max(created_at) f
from
(
    select uid, created_at, 
       conditional_true_event( created_at - lag(created_at) > '1 day' ) 
       over (partition by uid order by created_at) as seq_id
    from visits
) as seq
group by uid, seq_id;

最终产出：

uid  seq_id  num_days  s                    f                    
---  ------  --------  -------------------  -------------------  
123  0       3         2012-04-28 00:00:00  2012-04-30 00:00:00  
123  1       2         2012-05-03 00:00:00  2012-05-04 00:00:00  
123  2       2         2012-06-04 00:00:00  2012-06-04 00:00:00  

最后一点： num_days实际上是内部查询的行数。 如果原始表中有两个'2012-04-28'访问（即重复），您可能想要解决这个问题。

Answer 5

以下应该是Oracle友好的，不需要递归逻辑。

;WITH
  visit_dates (
    visit_id,
    date_id,
    group_id
  )
AS
(
  SELECT
    ROW_NUMBER() OVER (ORDER BY TRUNC(created_at)),
    TRUNC(SYSDATE) - TRUNC(created_at),
    TRUNC(SYSDATE) - TRUNC(created_at) - ROW_NUMBER() OVER (ORDER BY TRUNC(created_at))
  FROM
    visits
  GROUP BY
    TRUNC(created_at)
)
,
  group_duration (
    group_id,
    duration
  )
AS
(
  SELECT
    group_id,
    MAX(date_id) - MIN(date_id) + 1  AS duration
  FROM
    visit_dates
  GROUP BY
    group_id
)
SELECT
  MAX(duration)  AS max_duration
FROM
  group_duration

Answer 6

PostgreSQL的：

with headers as
(
    select 
        d,
        lag(d) over m is null or d - lag(d) over m  <> interval '1 day' as header

    from tbl
    window m as (order by d)
)      
,sequence_group as
(
    select d, sum(header::int) over m as group_number 
    from headers
    window m as (order by d)
)
,consecutive_list as
(
    select d, group_number, count(d) over m as consecutive_count
    from sequence_group 
    window m as (partition by group_number)
)
select * from consecutive_list

分而治之的方法：3个步骤

第一步，找到标题：

with headers as
(
    select 
        d,
        lag(d) over m is null or d - lag(d) over m  <> interval '1 day' as header

    from tbl
    window m as (order by d)
)
select * from headers

输出：

          d          | header 
---------------------+--------
 2012-04-28 08:00:00 | t
 2012-04-29 08:00:00 | f
 2012-04-30 08:00:00 | f
 2012-05-03 08:00:00 | t
 2012-05-04 08:00:00 | f
(5 rows)

第二步，指定分组：

with headers as
(
    select 
        d,
        lag(d) over m is null or d - lag(d) over m  <> interval '1 day' as header

    from tbl
    window m as (order by d)
)      
,sequence_group as
(
    select d, sum(header::int) over m as group_number 
    from headers
    window m as (order by d)
)
select * from sequence_group

输出：

          d          | group_number 
---------------------+--------------
 2012-04-28 08:00:00 |            1
 2012-04-29 08:00:00 |            1
 2012-04-30 08:00:00 |            1
 2012-05-03 08:00:00 |            2
 2012-05-04 08:00:00 |            2
(5 rows)

第3步，计算最大天数：

with headers as
(
    select 
        d,
        lag(d) over m is null or d - lag(d) over m  <> interval '1 day' as header

    from tbl
    window m as (order by d)
)      
,sequence_group as
(
    select d, sum(header::int) over m as group_number 
    from headers
    window m as (order by d)
)
,consecutive_list as
(
select d, group_number, count(d) over m as consecutive_count
from sequence_group 
window m as (partition by group_number)
)
select * from consecutive_list

输出：

          d          | group_number | consecutive_count 
---------------------+--------------+-----------------
 2012-04-28 08:00:00 |            1 |               3
 2012-04-29 08:00:00 |            1 |               3
 2012-04-30 08:00:00 |            1 |               3
 2012-05-03 08:00:00 |            2 |               2
 2012-05-04 08:00:00 |            2 |               2
(5 rows)

Answer 7

这是针对最短的MySQL，并使用最小变量（仅一个变量）：

select 
   min(d) as starting_date, max(d) as ending_date, 
   count(d) as consecutive_days
from
(
  select 
     sr.d,
     IF(fr.d is null,@group_number := @group_number + 1,@group_number) 
        as group_number
  from tbl sr
  left join tbl fr on sr.d = adddate(fr.d,interval 1 day)
  cross join (select @group_number := 0) as grp
) as x
group by group_number

输出：

STARTING_DATE                  ENDING_DATE                  CONSECUTIVE_DAYS
April, 28 2012 08:00:00-0700   April, 30 2012 08:00:00-0700 3
May, 03 2012 08:00:00-0700     May, 04 2012 08:00:00-0700   2

现场测试： http ： //www.sqlfiddle.com/#！2/65169/1

Answer 8

对于PostgreSQL 8.4或更高版本 ，有一个简短而干净的窗口函数方式，没有JOIN 。
我希望这是迄今为止发布的最快的解决方案 ：

WITH x AS (
    SELECT created_at AS d
         , lag(created_at) OVER (ORDER BY created_at) = (created_at - 1) AS nu
    FROM   visits
    WHERE  uid = 1
    )
   , y AS (
    SELECT d, count(NULLIF(nu, TRUE)) OVER (ORDER BY d) AS seq
    FROM   x
    )
SELECT count(*) AS max_days, min(d) AS seq_from,  max(d) AS seq_to
FROM   y
GROUP  BY seq
ORDER  BY 1 DESC
LIMIT  1;

返回：

max_days | seq_from   | seq_to
---------+------------+-----------
3        | 2012-04-28 | 2012-04-30

假设created_at是一个date且unique 。

1. 在CTE x：我们的用户每天访问，检查他是否也在昨天。 要计算“昨天”，只需使用created_at - 1第一行是特殊情况，并在此处生成NULL。

2. 在CTE y中：计算每天的“没有昨天的天数”（ seq ）的运行计数。 NULL值不计，所以count(NULLIF(nu, TRUE))是最简洁和最短的方式，也涵盖了特殊情况。

3. 最后，每组天seq和度日如年。 在此期间，我添加了序列的第一天和最后一天。 ORDER BY序列的长度，并选择最长的序列。

Answer 9

这个问题已有几个答案。 但是SQL语句看起来都太复杂了。 这可以通过基本SQL，枚举行的方法和一些日期算法来完成。

关键的观察结果是，如果你有一堆天并且有一个平行的整数序列，那么差异就是天数在一个序列中的一个恒定日期。

以下查询使用此观察来回答原始问题：

select uid, min(d) as startdate, count(*) as numdaysinseq
from 
(
   select uid, d, adddate(d, interval -offset day) as groupstart
   from 
   (
     select uid, d, row_number() over (partition by uid order by date) as offset
     from 
     (
       SELECT DISTINCT uid, DATE(created_at) AS d
       FROM visits
     ) t
   ) t
) t

唉，mysql没有row_number()函数。 但是，有一个变量的解决方法（大多数其他数据库都有这个功能）。

Answer 10

在看到OP的Vertica数据库查询方法后，我尝试同时运行两个连接：

这些Postgresql和Sql Server查询版本都适用于Vertica

Postgresql版本：

select 
  min(gr.d) as start_date,
  max(gr.d) as end_date,
  date_part('day', max(gr.d) - min(gr.d))+1 as consecutive_days
from 
(
  select 
  cr.d, (row_number() over() - 1) / 2 as pair_number
  from tbl cr   
  left join tbl pr on pr.d = cr.d - interval '1 day'
  left join tbl nr on nr.d = cr.d + interval '1 day'
  where pr.d is null <> nr.d is null
) as gr
group by pair_number
order by start_date

关于pr.d is null <> nr.d is null 。 这意味着，它是前一行为null或下一行为空，但它们永远不能都为空，所以这基本上删除了非连续日期，因为非连续日期的上一行和下一行是空的（这基本上给我们所有的日期只是页眉和页脚）。 这也称为XOR操作

如果我们只剩下连续日期，我们现在可以通过row_number配对它们：

(row_number() over() - 1) / 2 as pair_number

row_number()以1开头，我们需要用1减去它（我们也可以用1加减），然后我们将它除以2; 这使得配对日期彼此相邻

现场测试： http ： //www.sqlfiddle.com/#！1 / fc440 / 7

---

这是Sql Server版本：

select 
  min(gr.d) as start_date,
  max(gr.d) as end_date,
  datediff(day, min(gr.d),max(gr.d)) +1 as consecutive_days
from 
(
  select 
     cr.d, (row_number() over(order by cr.d) - 1) / 2 as pair_number
  from tbl cr   
  left join tbl pr on pr.d = dateadd(day,-1,cr.d)
  left join tbl nr on nr.d = dateadd(day,+1,cr.d)
  where         
       case when pr.d is null then 1 else 0 end
    <> case when nr.d is null then 1 else 0 end
) as gr
group by pair_number
order by start_date

与上述逻辑相同，除了日期函数的人为差异。 并且sql Server在其OVER上需要ORDER BY子句，而Postgresql的OVER可以保留为空。

Sql Server没有第一类布尔值，这就是我们无法直接比较布尔值的原因：

pr.d is null <> nr.d is null

我们必须在Sql Server中执行此操作：

   case when pr.d is null then 1 else 0 end
<> case when nr.d is null then 1 else 0 end

实时测试： http ： //www.sqlfiddle.com/#！3 / 65df2 / 17