我的架构如下（简化）：
CREATE TABLE MyTable (
DateTimeOffset HitDate NOT NULL,
IpAddress varchar(15)
)
示例行可能如下所示：
'7/10/2013 8:05:29 -07:00' '111.222.333.444'
我正在尝试找出一个查询，它会在每天（例如，2013年7月10日）为我提供唯一的IpAddresses数量。实际上，这部分相当简单，我已经为此创建了一个查询。但是，对于此查询，我想要的是在当前日期之前从未存在的唯一IpAddresses的数量。我不关心日期之后，就在日期之前。
例如，假设我有以下数据，这就是我所拥有的：
'7/10/2013 8:05:29 -07:00' '111.222.333.444'
'7/10/2013 12:05:29 -07:00' '111.222.333.222'
'7/9/2013 9:05:29 -07:00' '111.222.333.444'
'7/9/2013 10:05:29 -07:00' '111.222.333.555'
'7/8/2013 11:05:29 -07:00' '111.222.333.222'
'7/8/2013 4:05:29 -07:00' '111.222.333.555'
查询应输出以下内容：
'7/8/2013' 2 (neither IpAddress existed before this date so both are new)
'7/9/2013' 1 (only one of the IpAddresses is new - the one ending in '444')
'7/10/2013' 0 (both IpAddresses had existed before this date)
目标数据库为SQL Server 2012。我正在向第一个正确创建SQL语句的人提供100分的赏金。
答案 0 :(得分：3)
CREATE TABLE #MyTable (
HitDate DateTimeOffset NOT NULL,
IpAddress varchar(15))
insert #mytable values
('7/10/2013 8:05:29 -07:00', '111.222.333.444'),
('7/10/2013 12:05:29 -07:00', '111.222.333.222'),
('7/9/2013 9:05:29 -07:00' ,'111.222.333.444'),
('7/9/2013 10:05:29 -07:00', '111.222.333.555'),
('7/8/2013 11:05:29 -07:00', '111.222.333.222'),
('7/8/2013 4:05:29 -07:00', '111.222.333.555')
;WITH a AS
(
select
cast(HitDate as date) HitDate,
IpAddress
from #mytable
), b AS
(
SELECT min(HitDate) md, IpAddress
FROM a
GROUP BY IpAddress
)
SELECT c.HitDate, Count(distinct b.IpAddress) IpAddress
FROM b
right join (select distinct HitDate from a) c
on b.md = c.HitDate
GROUP by c.HitDate
结果：
HitDate IpAddress
2013-07-08 2
2013-07-09 1
2013-07-10 0
答案 1 :(得分：2)
尝试以下查询：
SELECT A.HITDATE , ISNULL(B.CNT , 0) AS COUNT
FROM
(
SELECT DISTINCT CONVERT(DATE , HITDATE) AS HITDATE
FROM mytable1
) AS A
LEFT OUTER JOIN
(
SELECT CONVERT(DATE , HITDATE) AS HITDATE , COUNT(*) AS CNT
FROM MyTable1 AS A
WHERE IpAddress NOT IN
(
SELECT B.IpAddress FROM mytable1 AS B WHERE A.HITDATE > B.HITDATE
)
GROUP BY CONVERT(DATE , HITDATE)
) AS B ON A.HITDATE = B.HITDATE
答案 2 :(得分：2)
这是我的解决方案：
CREATE TABLE Test(
DT datetime NOT NULL,
IP varchar(15)
)
INSERT INTO TEST
SELECT '2013-07-10', 'aaa' UNION ALL
SELECT '2013-07-10', 'bbb' UNION ALL
SELECT '2013-07-10', 'ccc' UNION ALL
SELECT '2013-07-09', 'aaa' UNION ALL
SELECT '2013-07-09', 'bbb' UNION ALL
SELECT '2013-07-08', 'aaa' UNION ALL
SELECT '2013-07-08', 'bbb'
SELECT CAST(T1.DT AS DATE) Date, COUNT(T1.IP)
FROM Test T1
WHERE T1.IP NOT IN (
SELECT DISTINCT T2.IP FROM Test T2 WHERE CAST(T2.DT AS DATE) < T1.DT
)
GROUP BY CAST(T1.DT AS DATE)
正确结果为：
Date: Unique IPs
2013-07-08 2
2013-07-10 1
答案 3 :(得分：2)
最简单的方法（在我看来）是找到IP地址出现的最早日期，然后将其用于聚合：
select cast(minHitDate as Date), count(*) as FirstTimeVisitors
from (select IpAddress, min(HitDate) as minHitDate
from MyTable t
group by IpAddress
) i
group by cast(minHitDate as Date)
order by 1;
另一种形式，可让您计算第一次访问者，第二次访问者等，使用dense_rank()：
select cast(HitDate as Date),
count(distinct IpAddress) as NumVisitors,
sum(case when nth = 1 then 1 else 0 end) as FirstTime,
sum(case when nth = 2 then 1 else 0 end) as SecondTime,
sum(case when nth = 3 then 1 else 0 end) as ThirdTime
from (select IpAddress,
dense_rank() over (partition by IpAddress
order by cast(HitDate as date)
) as nth
from MyTable t
) i
group by cast(HitDate as Date)
order by 1;
答案 4 :(得分：1)
[编辑 - 改为使用DateTimeOffset]
我想出了以下内容，首先创建基础MyTable：
-- Test data creation to match schema & examples
create TABLE MyTable (
HitDate DateTimeOffset NOT NULL,
IpAddress varchar(15)
)
insert into MyTable values ('7/10/2013 8:05:29 -07:00', '111.222.333.444')
insert into MyTable values ('7/10/2013 12:05:29 -07:00', '111.222.333.222')
insert into MyTable values ('7/9/2013 9:05:29 -07:00', '111.222.333.444')
insert into MyTable values ('7/9/2013 10:05:29 -07:00', '111.222.333.555')
insert into MyTable values ('7/8/2013 11:05:29 -07:00', '111.222.333.222')
insert into MyTable values ('7/8/2013 4:05:29 -07:00', '111.222.333.555')
-- actual solution starts here
create TABLE #MyTable (
HitDate date,
IpAddress varchar(15)
)
-- populate data into required format for main query
insert into #MyTable SELECT HitDate, IpAddress FROM MyTable
-- main query
select distinct convert(varchar(10), HitDate, 101) HitDate, ISNULL(dateCounts.counter, 0) NewIPAddresses
from #MyTable mainDates
left outer join
(
select main.HitDate dateValue, count(*) counter from
#MyTable main
left outer join #MyTable sub on main.IpAddress = sub.IpAddress and main.HitDate > sub.HitDate
where sub.IpAddress is null
group by main.HitDate
) dateCounts on dateCounts.dateValue = HitDate
给出：
HitDate NewIPAddresses
-------------- --------------
07/08/2013 2
07/09/2013 1
07/10/2013 0
答案 5 :(得分：1)
试试这个
SELECT
CONVERT(DATE,T1.[HitDate]) AS [HitDate]
,SUM(CASE WHEN T2.[IpAddress] IS NULL THEN 1 ELSE 0 END) AS IpAddress
FROM
YourTableName T1
LEFT JOIN
YourTableName T2 ON T1.IpAddress = T2.IpAddress
AND
T1.[HitDate] > T2.[HitDate]
GROUP BY CONVERT(DATE,T1.[HitDate])
<强>输出
HitDate IpAddress
2013-07-08 2
2013-07-09 1
2013-07-10 0
答案 6 :(得分：0)
这看起来也不错，但有点复杂......
SELECT DISTINCT(IPAddress)
FROM [MyTable]
WHERE CONVERT(DATE , HITDATE) <= '2013-07-09'
GROUP BY IPAddress
HAVING COUNT(*) = 1 AND MAX(CONVERT(DATE , HITDATE)) = '2013-07-09';
答案 7 :(得分：0)
我在T-SQL中使用了一个游标来实现所需的结果。
以下代码：
INSERT INTO MyTable(HitDate, IpAddress) values ('7/10/2013 8:05:29 -07:00', '111.222.333.444')
INSERT INTO MyTable(HitDate, IpAddress) VALUES('7/10/2013 12:05:29 -07:00', '111.222.333.222')
INSERT INTO MyTable(HitDate, IpAddress) VALUES('7/9/2013 9:05:29 -07:00', '111.222.333.444')
INSERT INTO MyTable(HitDate, IpAddress) VALUES('7/9/2013 10:05:29 -07:00', '111.222.333.555')
INSERT INTO MyTable(HitDate, IpAddress) VALUES('7/8/2013 11:05:29 -07:00', '111.222.333.222')
INSERT INTO MyTable(HitDate, IpAddress) VALUES('7/8/2013 4:05:29 -07:00', '111.222.333.555')
DECLARE @HitDate DATE
DECLARE cCursor CURSOR FAST_FORWARD LOCAL
FOR
SELECT
DISTINCT CAST(HitDate AS DATE)
FROM
MyTable
ORDER BY
CAST(HitDate AS DATE) ASC
OPEN cCursor
FETCH NEXT FROM cCursor INTO @HitDate
WHILE (@@FETCH_STATUS = 0)
BEGIN
SELECT @HitDate, COUNT(IpAddress)
FROM MyTable
WHERE IpAddress NOT IN (SELECT IpAddress FROM MyTable WHERE HitDate < @HitDate) AND CAST(HitDate AS DATE) = @HitDate
FETCH NEXT FROM cCursor INTO @HitDate
END
CLOSE cCursor
DEALLOCATE cCursor
输出如下：
2013-07-08 2
2013-07-09 1
2013-07-10 0
答案 8 :(得分：0)
试试这个：
SELECT CONVERT(DATE, mt.HitDate) AS hitDate
, COUNT(firstDate.ipAddress) AS newIPAddresses
FROM MyTable mt
LEFT JOIN (SELECT IpAddress, MIN(HitDate) AS FirstHitDate
FROM MyTable
GROUP BY IpAddress) firstDate
ON mt.HitDate = firstDate.FirstHitDate
AND mt.IpAddress = firstDate.IpAddress
GROUP BY CONVERT(DATE, mt.HitDate)
ORDER BY 1
输出：
hitDate newIPAddress
2013-07-08 2
2013-07-09 1
2013-07-10 0