4

我有一个如下所示的架构（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'

我正在尝试制定一个查询，该查询将为我提供每天（例如，2013 年 7 月 10 日）唯一 IpAddress 的数量。实际上，这部分相当简单，我已经为此创建了一个查询。但是，对于这个查询，我想要的是IpAddresses在当前日期之前从未存在过的唯一数量。我不在乎日期之后，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)

目标数据库是SQL Server 2012. 我向第一个正确创建 SQL 语句的人提供 100 分的奖励。

4

9 回答 9

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
于 2013-07-10T16:01:13.117 回答
2

This is my solution:

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)

which correctly results as:

Date:       Unique IPs
2013-07-08  2
2013-07-10  1
于 2013-07-10T12:41:31.773 回答
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

http://sqlfiddle.com/#!6/7536a/1

于 2013-07-10T12:35:42.733 回答
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;

可让您计算第一次访问者、第二次访问者等的另一种形式使用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;
于 2013-07-13T18:52:44.393 回答
1

尝试这个

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])

SQL 小提琴演示

data.stackexchange 演示

输出

HitDate    IpAddress
2013-07-08  2
2013-07-09  1
2013-07-10  0
于 2013-07-15T05:58:34.830 回答
1

[编辑 - 改为使用 DateTimeOffset]

我想出了以下内容，从创建基础 MyTable 开始：

-- 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

这使：

    HitDate        NewIPAddresses 
    -------------- --------------
    07/08/2013     2
    07/09/2013     1
    07/10/2013     0
于 2013-07-10T12:27:59.003 回答
0

尝试这个：

    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
于 2013-07-15T05:38:34.570 回答
0

我在 T-SQL 中使用了一个游标来实现所需的结果。

下面的代码：

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

于 2013-07-15T04:35:45.157 回答
0

这看起来也不错，但不那么复杂......

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';
于 2013-07-10T13:16:05.613 回答