SQL确定最少连续访问天数？

以下用户历史记录表包含给定用户每天（在UTC的24小时内）访问网站的每一条记录。它有成千上万条记录，但每个用户每天只有一条记录。如果用户当天没有访问该网站，则不会生成任何记录。

ID UserId CreationDate
------ ------ ------------
750997 12 2009-07-07 18：42：20.723
750998 15 2009-07-07 18：42：20.927
751000 19 2009-07-07 18：42：22.283

我正在寻找的是对此表具有良好性能的SQL查询，该查询告诉我哪些用户ID连续（n）天访问了该网站，却没有一天丢失。

换句话说，该表中有（n）个记录具有连续（前一天或后一天）日期的记录？如果序列中缺少任何一天，则该序列将中断并且应从1重新开始；否则，序列将重新开始。我们正在寻找在此连续几天没有间断的用户。

当然，此查询与特定的Stack Overflow标志之间的任何相似之处纯属巧合。.：:)

答案显然是：

SELECT DISTINCT UserId
FROM UserHistory uh1
WHERE (
       SELECT COUNT(*) 
       FROM UserHistory uh2 
       WHERE uh2.CreationDate 
       BETWEEN uh1.CreationDate AND DATEADD(d, @days, uh1.CreationDate)
      ) = @days OR UserId = 52551

编辑：

好吧，这是我的认真回答：

DECLARE @days int
DECLARE @seconds bigint
SET @days = 30
SET @seconds = (@days * 24 * 60 * 60) - 1
SELECT DISTINCT UserId
FROM (
    SELECT uh1.UserId, Count(uh1.Id) as Conseq
    FROM UserHistory uh1
    INNER JOIN UserHistory uh2 ON uh2.CreationDate 
        BETWEEN uh1.CreationDate AND 
            DATEADD(s, @seconds, DATEADD(dd, DATEDIFF(dd, 0, uh1.CreationDate), 0))
        AND uh1.UserId = uh2.UserId
    GROUP BY uh1.Id, uh1.UserId
    ) as Tbl
WHERE Conseq >= @days

编辑：

[Jeff Atwood]这是一个很棒的快速解决方案，值得接受，但是Rob Farley的解决方案也非常出色，而且可以说甚至更快（！）。请也检查一下！

怎么样（请确保前面的语句以分号结尾）：

WITH numberedrows
     AS (SELECT ROW_NUMBER() OVER (PARTITION BY UserID 
                                       ORDER BY CreationDate)
                - DATEDIFF(day,'19000101',CreationDate) AS TheOffset,
                CreationDate,
                UserID
         FROM   tablename)
SELECT MIN(CreationDate),
       MAX(CreationDate),
       COUNT(*) AS NumConsecutiveDays,
       UserID
FROM   numberedrows
GROUP  BY UserID,
          TheOffset  

这样的想法是，如果我们拥有日期列表（作为数字）和row_number，那么错过的日期会使这两个列表之间的偏移量稍大。因此，我们正在寻找具有一致偏移的范围。

您可以在此末尾使用“ ORDER BY NumConsecutiveDays DESC”，或说“ HAVING count（*）> 14”作为阈值...

我还没有测试过-只是把它写在我的头上。希望可以在SQL2005及更高版本上使用。

...并且通过表名的索引（UserID，CreationDate）将大有帮助

编辑：原来偏移是保留字，所以我改用TheOffset。

编辑：建议使用COUNT（*）是非常有效的-首先我应该这样做，但实际上并没有考虑。以前它使用的是datediff（day，min（CreationDate），max（CreationDate））。

抢

如果您可以更改表格架构，建议您在表格中添加一列LongestStreak，并将其设置为以结尾的连续天数CreationDate。在登录时更新表很容易（类似于您已经在做的事情，如果当天没有行，您将检查前一天是否有任何行。如果为true，则LongestStreak在新行，否则，将其设置为1。）

添加此列后，查询将显而易见：

if exists(select * from table
          where LongestStreak >= 30 and UserId = @UserId)
   -- award the Woot badge.

一些表现力很好的SQL，大致表现为：

select
        userId,
    dbo.MaxConsecutiveDates(CreationDate) as blah
from
    dbo.Logins
group by
    userId

假设您有一个用户定义的聚合函数，大致类似（请注意，这是错误的）：

using System;
using System.Data.SqlTypes;
using Microsoft.SqlServer.Server;
using System.Runtime.InteropServices;

namespace SqlServerProject1
{
    [StructLayout(LayoutKind.Sequential)]
    [Serializable]
    internal struct MaxConsecutiveState
    {
        public int CurrentSequentialDays;
        public int MaxSequentialDays;
        public SqlDateTime LastDate;
    }

    [Serializable]
    [SqlUserDefinedAggregate(
        Format.Native,
        IsInvariantToNulls = true, //optimizer property
        IsInvariantToDuplicates = false, //optimizer property
        IsInvariantToOrder = false) //optimizer property
    ]
    [StructLayout(LayoutKind.Sequential)]
    public class MaxConsecutiveDates
    {
        /// <summary>
        /// The variable that holds the intermediate result of the concatenation
        /// </summary>
        private MaxConsecutiveState _intermediateResult;

        /// <summary>
        /// Initialize the internal data structures
        /// </summary>
        public void Init()
        {
            _intermediateResult = new MaxConsecutiveState { LastDate = SqlDateTime.MinValue, CurrentSequentialDays = 0, MaxSequentialDays = 0 };
        }

        /// <summary>
        /// Accumulate the next value, not if the value is null
        /// </summary>
        /// <param name="value"></param>
        public void Accumulate(SqlDateTime value)
        {
            if (value.IsNull)
            {
                return;
            }
            int sequentialDays = _intermediateResult.CurrentSequentialDays;
            int maxSequentialDays = _intermediateResult.MaxSequentialDays;
            DateTime currentDate = value.Value.Date;
            if (currentDate.AddDays(-1).Equals(new DateTime(_intermediateResult.LastDate.TimeTicks)))
                sequentialDays++;
            else
            {
                maxSequentialDays = Math.Max(sequentialDays, maxSequentialDays);
                sequentialDays = 1;
            }
            _intermediateResult = new MaxConsecutiveState
                                      {
                                          CurrentSequentialDays = sequentialDays,
                                          LastDate = currentDate,
                                          MaxSequentialDays = maxSequentialDays
                                      };
        }

        /// <summary>
        /// Merge the partially computed aggregate with this aggregate.
        /// </summary>
        /// <param name="other"></param>
        public void Merge(MaxConsecutiveDates other)
        {
            // add stuff for two separate calculations
        }

        /// <summary>
        /// Called at the end of aggregation, to return the results of the aggregation.
        /// </summary>
        /// <returns></returns>
        public SqlInt32 Terminate()
        {
            int max = Math.Max((int) ((sbyte) _intermediateResult.CurrentSequentialDays), (sbyte) _intermediateResult.MaxSequentialDays);
            return new SqlInt32(max);
        }
    }
}

似乎您可以利用以下事实：要连续n天，则需要有n行。

所以像这样：

SELECT users.UserId, count(1) as cnt
FROM users
WHERE users.CreationDate > now() - INTERVAL 30 DAY
GROUP BY UserId
HAVING cnt = 30

对我而言，使用单个SQL查询执行此操作似乎过于复杂。让我将这个答案分为两个部分。

1. 您应该做的到现在为止，现在应该开始做：
   运行一项日常cron作业，检查他今天登录的每个用户，然后增加一个计数器（如果有）或将其设置为0（如果没有）。
2. 您现在应该执行的操作：
   -将该表导出到不会运行您的网站且一会儿不需要的服务器。;）
   -按用户排序，然后按日期排序。
   -依次浏览，保持计数器...

如果这对您来说如此重要，请发送该事件并驱动一个表以提供此信息。无需使用所有这些疯狂的查询来杀死机器。

您可以使用递归CTE（SQL Server 2005+）：

WITH recur_date AS (
        SELECT t.userid,
               t.creationDate,
               DATEADD(day, 1, t.created) 'nextDay',
               1 'level' 
          FROM TABLE t
         UNION ALL
        SELECT t.userid,
               t.creationDate,
               DATEADD(day, 1, t.created) 'nextDay',
               rd.level + 1 'level'
          FROM TABLE t
          JOIN recur_date rd on t.creationDate = rd.nextDay AND t.userid = rd.userid)
   SELECT t.*
    FROM recur_date t
   WHERE t.level = @numDays
ORDER BY t.userid

Joe Celko在有关Smarties的SQL中对此有完整的章节（称为运行和序列）。我家里没有那本书，所以当我上班时……我会回答的。（假设历史记录表称为dbo.UserHistory，天数为@Days）

另一个线索来自SQL Team的运行博客

我有另一个想法，但是这里没有方便使用的SQL Server是将CTE与分区ROW_NUMBER一起使用，如下所示：

WITH Runs
AS
  (SELECT UserID
         , CreationDate
         , ROW_NUMBER() OVER(PARTITION BY UserId
                             ORDER BY CreationDate)
           - ROW_NUMBER() OVER(PARTITION BY UserId, NoBreak
                               ORDER BY CreationDate) AS RunNumber
  FROM
     (SELECT UH.UserID
           , UH.CreationDate
           , ISNULL((SELECT TOP 1 1 
              FROM dbo.UserHistory AS Prior 
              WHERE Prior.UserId = UH.UserId 
              AND Prior.CreationDate
                  BETWEEN DATEADD(dd, DATEDIFF(dd, 0, UH.CreationDate), -1)
                  AND DATEADD(dd, DATEDIFF(dd, 0, UH.CreationDate), 0)), 0) AS NoBreak
      FROM dbo.UserHistory AS UH) AS Consecutive
)
SELECT UserID, MIN(CreationDate) AS RunStart, MAX(CreationDate) AS RunEnd
FROM Runs
GROUP BY UserID, RunNumber
HAVING DATEDIFF(dd, MIN(CreationDate), MAX(CreationDate)) >= @Days

上面的内容可能比必须的要困难得多，但是当您对“运行”有除日期以外的其他定义时，上面的内容会让人脑子发痒。

几个SQL Server 2012选项（假设下面的N = 100）。

;WITH T(UserID, NRowsPrevious)
     AS (SELECT UserID,
                DATEDIFF(DAY, 
                        LAG(CreationDate, 100) 
                            OVER 
                                (PARTITION BY UserID 
                                     ORDER BY CreationDate), 
                         CreationDate)
         FROM   UserHistory)
SELECT DISTINCT UserID
FROM   T
WHERE  NRowsPrevious = 100 

尽管使用我的样本数据，以下结果更加有效

;WITH U
         AS (SELECT DISTINCT UserId
             FROM   UserHistory) /*Ideally replace with Users table*/
    SELECT UserId
    FROM   U
           CROSS APPLY (SELECT TOP 1 *
                        FROM   (SELECT 
                                       DATEDIFF(DAY, 
                                                LAG(CreationDate, 100) 
                                                  OVER 
                                                   (ORDER BY CreationDate), 
                                                 CreationDate)
                                FROM   UserHistory UH
                                WHERE  U.UserId = UH.UserID) T(NRowsPrevious)
                        WHERE  NRowsPrevious = 100) O

两者都依赖于该问题中所述的约束条件，即每个用户每天最多有一个记录。

像这样吗

select distinct userid
from table t1, table t2
where t1.UserId = t2.UserId 
  AND trunc(t1.CreationDate) = trunc(t2.CreationDate) + n
  AND (
    select count(*)
    from table t3
    where t1.UserId  = t3.UserId
      and CreationDate between trunc(t1.CreationDate) and trunc(t1.CreationDate)+n
   ) = n

我使用一个简单的数学属性来确定谁连续访问了该网站。此属性是，您首次访问和最后一次访问之间的日差应等于访问表日志中的记录数。

这是我在Oracle DB中测试过的SQL脚本（它也应该在其他DB中工作）：

-- show basic understand of the math properties 
  select    ceil(max (creation_date) - min (creation_date))
              max_min_days_diff,
           count ( * ) real_day_count
    from   user_access_log
group by   user_id;


-- select all users that have consecutively accessed the site 
  select   user_id
    from   user_access_log
group by   user_id
  having       ceil(max (creation_date) - min (creation_date))
           / count ( * ) = 1;



-- get the count of all users that have consecutively accessed the site 
  select   count(user_id) user_count
    from   user_access_log
group by   user_id
  having   ceil(max (creation_date) - min (creation_date))
           / count ( * ) = 1;

表准备脚本：

-- create table 
create table user_access_log (id           number, user_id      number, creation_date date);


-- insert seed data 
insert into user_access_log (id, user_id, creation_date)
  values   (1, 12, sysdate);

insert into user_access_log (id, user_id, creation_date)
  values   (2, 12, sysdate + 1);

insert into user_access_log (id, user_id, creation_date)
  values   (3, 12, sysdate + 2);

insert into user_access_log (id, user_id, creation_date)
  values   (4, 16, sysdate);

insert into user_access_log (id, user_id, creation_date)
  values   (5, 16, sysdate + 1);

insert into user_access_log (id, user_id, creation_date)
  values   (6, 16, sysdate + 5);

declare @startdate as datetime, @days as int
set @startdate = cast('11 Jan 2009' as datetime) -- The startdate
set @days = 5 -- The number of consecutive days

SELECT userid
      ,count(1) as [Number of Consecutive Days]
FROM UserHistory
WHERE creationdate >= @startdate
AND creationdate < dateadd(dd, @days, cast(convert(char(11), @startdate, 113)  as datetime))
GROUP BY userid
HAVING count(1) >= @days

该声明 cast(convert(char(11), @startdate, 113) as datetime)删除了日期的时间部分，因此我们从午夜开始。

我也假设creationdate和userid列都已建立索引。

我只是意识到，这不会告诉您所有用户及其连续的总天数。但是会告诉您，从您选择的日期开始，哪些用户已经访问了指定的天数。

修改后的解决方案：

declare @days as int
set @days = 30
select t1.userid
from UserHistory t1
where (select count(1) 
       from UserHistory t3 
       where t3.userid = t1.userid
       and t3.creationdate >= DATEADD(dd, DATEDIFF(dd, 0, t1.creationdate), 0) 
       and t3.creationdate < DATEADD(dd, DATEDIFF(dd, 0, t1.creationdate) + @days, 0) 
       group by t3.userid
) >= @days
group by t1.userid

我已经检查了它，它将查询所有用户和所有日期。它基于Spencer的第一个（笑话？）解决方案，但是我的工作正常。

更新：改进了第二个解决方案中的日期处理。

这应该可以满足您的要求，但是我没有足够的数据来测试效率。复杂的CONVERT / FLOOR内容是将时间部分从datetime字段中剥离。如果您使用的是SQL Server 2008，则可以使用CAST（x.CreationDate AS DATE）。

声明@Range为INT
SET @范围= 10

SELECT DISTINCT UserId，CONVERT（DATETIME，FLOOR（CONVERT（FLOAT，a.CreationDate）））
  从tblUserLogin一个
存在的地方
   （选择1 
      来自tblUserLogin b 
     其中a.userId = b.userId 
       AND（SELECT COUNT（DISTINCT（CONVERT（DATETIME，FLOOR（CONVERT（FLOAT，CreationDate）））））） 
              来自tblUserLogin c 
             在哪里c.userid = b.userid 
               AND CONVERT（DATETIME，FLOOR（CONVERT（FLOAT，c.CreationDate）））在CONVERT（DATETIME，FLOOR（CONVERT（FLOAT，a.CreationDate）））和CONVERT（DATETIME，FLOOR（CONVERT（FLOAT，a.CreationDate）））之间）+ @ Range-1）= @Range）

创建脚本

创建表[dbo]。[tblUserLogin]（
    [Id] [int] IDENTITY（1,1）NOT NULL，
    [UserId] [int] NULL，
    [CreationDate] [datetime] NULL
）在[PRIMARY]上

Spencer几乎做到了，但这应该是工作代码：

SELECT DISTINCT UserId
FROM History h1
WHERE (
    SELECT COUNT(*) 
    FROM History
    WHERE UserId = h1.UserId AND CreationDate BETWEEN h1.CreationDate AND DATEADD(d, @n-1, h1.CreationDate)
) >= @n

在我的头上，MySQLish：

SELECT start.UserId
FROM UserHistory AS start
  LEFT OUTER JOIN UserHistory AS pre_start ON pre_start.UserId=start.UserId
    AND DATE(pre_start.CreationDate)=DATE_SUB(DATE(start.CreationDate), INTERVAL 1 DAY)
  LEFT OUTER JOIN UserHistory AS subsequent ON subsequent.UserId=start.UserId
    AND DATE(subsequent.CreationDate)<=DATE_ADD(DATE(start.CreationDate), INTERVAL 30 DAY)
WHERE pre_start.Id IS NULL
GROUP BY start.Id
HAVING COUNT(subsequent.Id)=30

未经测试，几乎可以肯定，MSSQL需要进行一些转换，但是我认为可以提供一些想法。

如何使用Tally表？它遵循一种更加算法化的方法，执行计划轻而易举。在tallyTable中填充要扫描的表中从1到'MaxDaysBehind'的数字（即90将查找3个月，依此类推）。

declare @ContinousDays int
set @ContinousDays = 30  -- select those that have 30 consecutive days

create table #tallyTable (Tally int)
insert into #tallyTable values (1)
...
insert into #tallyTable values (90) -- insert numbers for as many days behind as you want to scan

select [UserId],count(*),t.Tally from HistoryTable 
join #tallyTable as t on t.Tally>0
where [CreationDate]> getdate()-@ContinousDays-t.Tally and 
      [CreationDate]<getdate()-t.Tally 
group by [UserId],t.Tally 
having count(*)>=@ContinousDays

delete #tallyTable

稍微调整Bill的查询。您可能需要在分组之前截断日期，以便每天只计算一次登录...

SELECT UserId from History 
WHERE CreationDate > ( now() - n )
GROUP BY UserId, 
DATEADD(dd, DATEDIFF(dd, 0, CreationDate), 0) AS TruncatedCreationDate  
HAVING COUNT(TruncatedCreationDate) >= n

编辑以使用DATEADD（dd，DATEDIFF（dd，0，CreationDate），0）代替convert（char（10），CreationDate，101）。

@IDisposable我想早点使用datepart，但是我懒得查找语法，所以我发现id改用convert。我知道这会产生重大影响谢谢！现在我明白了。

假设模式如下：

create table dba.visits
(
    id  integer not null,
    user_id integer not null,
    creation_date date not null
);

这将从带有间隔的日期序列中提取连续范围。

select l.creation_date  as start_d, -- Get first date in contiguous range
    (
        select min(a.creation_date ) as creation_date 
        from "DBA"."visits" a 
            left outer join "DBA"."visits" b on 
                   a.creation_date = dateadd(day, -1, b.creation_date ) and 
                   a.user_id  = b.user_id 
            where b.creation_date  is null and
                  a.creation_date  >= l.creation_date  and
                  a.user_id  = l.user_id 
    ) as end_d -- Get last date in contiguous range
from  "DBA"."visits" l
    left outer join "DBA"."visits" r on 
        r.creation_date  = dateadd(day, -1, l.creation_date ) and 
        r.user_id  = l.user_id 
    where r.creation_date  is null