sql - SQL 问题 - 计算最大天数序列

Question

有一张包含访问数据的表格：

uid (INT) | created_at (DATETIME)

我想知道用户连续多少天访问了我们的应用程序。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 条记录和两个间隔 - 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该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/)，但是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。

好吧，大多数解决方案都需要稍作修改。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

另一种最短的方法是进行自联接：

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（或者前一行？ツ</a>）。基本上我们正在做一个回溯，这是一个不支持的数据库模拟滞后LAG（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，我们使用了一种混合方法，通过连接模拟 LAG，然后对它使用 SUM 窗口，这会产生组号

更新

忘了放最后一个查询，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

更新

我知道为什么我使用窗口函数的其他解决方案变得很长，因为我试图说明组编号和对组计数的逻辑而变得很长。如果我像MySql 方法那样切入正题，那么窗口函数可能会更短。话虽如此，这是我的旧窗口函数方法，尽管现在更好：

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，您可以这样做：

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，我不能 100% 确定这是最好的方法。

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 有一些类似于 MSSQL 中可用的公用表表达式的东西。我对 Postgresql 不是很熟悉，但是下面的代码适用于 MSSQL 并且可以满足您的需求。

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来做出贡献（毕竟它是您支付的一部分）。我保证最后的查询很短。

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

为每一行分配一个事件窗口编号，从 0 开始，并在布尔参数表达式的结果为真时将编号增加 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     

现在最终查询变得简单：

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;

最终输出：

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。如果原始表中有两次'2012-04-28'访问（即重复），您可能需要解决这个问题。

Answer 5

这适用于最短的 MySQL，并使用最少的变量（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 6

对于PostgreSQL 8.4 或更高版本，有一个简短而干净的方法，带有窗口函数，而没有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;

回报：

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

假设这created_at是一个date和unique。

1. 在 CTE x 中：对于我们的用户访问的每一天，检查他昨天是否也在这里。要计算“昨天”，只需使用created_at - 1第一行是一种特殊情况，在这里会产生 NULL。

2. seq在 CTE y 中：计算每天的“到目前为止没有昨天的天数”（ ）的运行计数。NULL 值不计算在内，count(NULLIF(nu, TRUE))最快和最短的方法也是如此，也涵盖了特殊情况。

3. 最后，分组天数seq并计算天数。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，并选择最长的一个。

Answer 7

以下应该是 Oracle 友好的，并且不需要递归逻辑。

;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 8

PostgreSQL：

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)

第三步，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 9

在看到 OP 对其 Vertica 数据库的查询方法后，我尝试让两个连接同时运行：

这些 Postgresql 和 Sql Server 查询版本都应在 Vertica 中工作

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，前一行为空或下一行为空，但它们永远不能同时为空，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为空（这基本上给我们所有只是页眉和页脚的日期）。这也称为XOR 操作

如果我们只剩下连续的日期，我们现在可以通过 row_number 将它们配对：

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

row_number()从 1 开始，我们需要用 1 减去它（我们也可以用 1 代替），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

与上述相同的逻辑，除了日期函数的人为差异。并且 sql Server 需要一个ORDER BY子句OVER，而 PostgresqlOVER可以留空。

Sql Server 没有一流的布尔值，这就是我们不能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 中这样做：

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

现场测试：http ://www.sqlfiddle.com/#!3/65df2/17

Answer 10

这个问题已经有好几个答案了。然而，SQL 语句似乎都太复杂了。这可以通过基本的 SQL、枚举行的方法和一些日期算术来完成。

关键的观察是，如果你有一堆天并且有一个平行的整数序列，那么当这些天在一个序列中时，差异就是一个恒定的日期。

以下查询使用此观察结果来回答原始问题：

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没有这个row_number()功能。但是，有一个变量的变通方法（大多数其他数据库确实有这个功能）。