以下用户历史记录表包含给定用户访问网站的每一天的一条记录（在 24 小时 UTC 时间段内）。它有数千条记录，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
我正在寻找的是该表上具有良好性能的 SQL 查询，它告诉我哪些用户 ID 连续 (n) 天访问了该网站而没有错过任何一天。
换句话说，有多少用户在此表中有 (n) 条记录，其日期是连续的（前天或后天）？如果序列中缺少任何一天，则序列中断，应从 1 重新开始；我们正在寻找已经在此连续使用天数且没有间隔的用户。
当然，此查询与特定 Stack Overflow 标志之间的任何相似之处纯属巧合.. :)
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
这个想法是，如果我们有天数列表（作为一个数字）和一个 row_number，那么错过的天数会使这两个列表之间的偏移量稍微大一些。所以我们正在寻找一个具有一致偏移量的范围。
您可以在结尾处使用“ORDER BY NumConsecutiveDays DESC”，或者说“HAVING count(*) > 14”作为阈值...
不过，我还没有对此进行测试-只是将其从头顶写下来。希望适用于 SQL2005 及更高版本。
...并且会通过 tablename(UserID, CreationDate) 上的索引得到很大帮助
编辑：原来 Offset 是一个保留字，所以我改用了 TheOffset 。
已编辑：使用 COUNT(*) 的建议非常有效——我一开始就应该这样做，但并没有真正考虑。以前它使用 datediff(day, min(CreationDate), max(CreationDate)) 代替。
抢
答案很明显：
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：
好的，这是我的严肃回答：
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，值得被接受，但Rob Farley 的解决方案也非常出色，甚至可以说更快（！）。请也检查一下！
如果您可以更改表架构，我建议LongestStreak您在表中添加一列，将其设置为以CreationDate. 在登录时更新表很容易（类似于您已经在做的事情，如果当天不存在任何行，您将检查前一天是否存在任何行。如果为真，您将LongestStreak增加新行，否则，您将其设置为 1。）
添加此列后查询将很明显：
if exists(select * from table
where LongestStreak >= 30 and UserId = @UserId)
-- award the Woot badge.
一些很好表达的 SQL，如下所示：
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（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);
}
}
}
似乎您可以利用这样一个事实，即连续 n 天需要有 n 行。
所以像：
SELECT users.UserId, count(1) as cnt
FROM users
WHERE users.CreationDate > now() - INTERVAL 30 DAY
GROUP BY UserId
HAVING cnt = 30
对我来说，使用单个 SQL 查询执行此操作似乎过于复杂。让我把这个答案分成两部分。
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 SQL for Smarties 中有一个完整的章节（称为运行和序列）。我家里没有那本书，所以当我上班时......我会真正回答这个问题。（假设历史表称为 dbo.UserHistory 并且天数是 @Days）
另一个线索来自SQL 团队关于运行的博客
我有另一个想法，但在这里没有方便的 SQL 服务器可以使用，是使用带有分区 ROW_NUMBER 的 CTE，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
上面的内容可能比它必须的要难得多，但是当你对“跑步”有其他定义时，不仅仅是约会。
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
尽管使用我的示例数据，以下结果更有效
;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
两者都依赖于问题中所述的约束，即每个用户每天最多有一条记录。
如果这对您来说如此重要，请获取此事件并驱动表格为您提供此信息。无需用所有这些疯狂的查询来杀死机器。
我使用了一个简单的数学属性来确定谁连续访问了该站点。此属性是您应该使第一次访问和最后一次之间的天差等于访问表日志中的记录数。
这是我在 Oracle DB 中测试的 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);
像这样的东西？
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
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
该语句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。
我还假设creationdate和userid列已编入索引。
我刚刚意识到这不会告诉您所有用户及其连续天数。但会告诉您从您选择的日期起，哪些用户将访问固定天数。
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
我已经检查过了，它将查询所有用户和所有日期。它基于斯宾塞的第一个（笑话？）解决方案，但我的作品。
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，但我没有足够的数据来测试效率。令人费解的 CONVERT/FLOOR 内容是从日期时间字段中剥离时间部分。如果您使用的是 SQL Server 2008，那么您可以使用 CAST(x.CreationDate AS DATE)。
将@Range 声明为 INT
设置@范围 = 10
SELECT DISTINCT UserId, CONVERT(DATETIME, FLOOR(CONVERT(FLOAT, a.CreationDate)))
FROM tblUserLogin a
存在于何处
（选择 1
FROM tblUserLogin b
其中 a.userId = b.userId
AND (SELECT COUNT(DISTINCT(CONVERT(DATETIME, FLOOR(CONVERT(FLOAT, CreationDate)))))
FROM tblUserLogin c
其中 c.userid = b.userid
AND CONVERT(DATETIME, FLOOR(CONVERT(FLOAT, c.CreationDate))) CONVERT(DATETIME, FLOOR(CONVERT(FLOAT, a.CreationDate))) 和 CONVERT(DATETIME, FLOOR(CONVERT(FLOAT, a.CreationDate)) )+@Range-1) = @Range)
创建脚本
创建表 [dbo].[tblUserLogin](
[Id] [int] IDENTITY(1,1) 非空，
[用户 ID] [int] NULL，
[创建日期] [日期时间] NULL
) 开 [主要]
稍微调整比尔的查询。您可能必须在分组之前截断日期以仅计算每天一次登录...
SELECT UserId from History
WHERE CreationDate > ( now() - n )
GROUP BY UserId,
DATEADD(dd, DATEDIFF(dd, 0, CreationDate), 0) AS TruncatedCreationDate
HAVING COUNT(TruncatedCreationDate) >= n
编辑使用 DATEADD(dd, DATEDIFF(dd, 0, CreationDate), 0) 而不是 convert( char(10) , CreationDate, 101 )。
@IDisposable 我之前想使用 datepart 但我懒得查找语法，所以我想 id 使用 convert 代替。我不知道它产生了重大影响 谢谢！现在我知道。
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
在我的脑海中，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，几乎可以肯定需要对 MSSQL 进行一些转换，但我认为这给出了一些想法。
使用 Tally 表怎么样？它遵循更算法化的方法，执行计划轻而易举。使用从 1 到“MaxDaysBehind”之间的数字填充您要扫描的表格（即 90 将在 3 个月后查找等）。
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
假设一个模式是这样的：
create table dba.visits
(
id integer not null,
user_id integer not null,
creation_date date not null
);
这将从具有间隙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