0

我们想要一个 SQL 语句，列出唯一 IP/唯一 ID 对在任何唯一日期访问的次数，按唯一 ID/IP 对访问的最大次数排序。

这是表结构：

Column             Type
------------------------------
Date               Timestamp
NumberofUsers      smallint
ipaddress          varchar(16)
location           varchar(2)
Count              bigint(20)

这是我们一直在尝试的sql：

SELECT
  LicenseID, 
  MAX(Date) AS LatestAccess,
  COUNT(DISTINCT Location) AS DifferentCountries,
  COUNT(DISTINCT IPAddress) AS DistinctIPCount,
  COUNT(DISTINCT Date,IPAddress) AS DistinctDate
FROM
  LicenseHistory 
WHERE
  (LicenseID<>30002)
GROUP BY
  LicenseID 
ORDER BY
  DistinctDate DESC

以下是表格中 CSV 格式的一些示例日期：

2009-10-08 10:37,30002,8,24.108.64.80,CA,2399
2009-05-27 16:57,24508,50,24.108.64.80,CA,645
2008-11-06 12:04,30,100,24.108.64.80,CA,282
2008-02-04 10:51,24508,30,24.69.19.207,CA,62
2009-10-08 14:52,13136,5,24.108.64.80,CA,285
2013-05-13 13:10,718,10,66.251.68.106,US,23860
2008-02-12 11:10,30002,8,24.69.19.207,CA,36
2008-04-09 17:49,18504,10,70.90.32.57,US,121
2007-07-26 13:38,30002,8,76.226.201.191,US,2
2009-12-03 22:35,30002,8,196.25.255.214,ZA,14
2013-05-13 6:49,20341,4,66.232.201.125,US,2676
2007-07-28 23:57,30002,8,75.81.107.238,US,1
2007-07-29 10:39,30002,8,70.63.54.162,US,1
2007-07-30 3:53,30002,8,121.210.199.31,AU,4
2007-07-30 5:11,30002,8,41.207.67.10,KE,2

这是一些示例结果（尚不正确，最后一列不应与倒数第二个匹配）：

uniqueID  LatestAccess         DifferentCountries  DistinctIPCount  DistinctDate
--------------------------------------------------------------------------------
20677     2013-05-13 18:20:15  4                   162              162
27749     2013-05-14 05:30:59  7                   155              155
459       2013-05-13 11:12:47  2                   143              143
24965     2013-05-14 13:44:56  6                   123              123
25226     2013-05-06 16:11:56  3                   104              104
20370     2013-05-14 05:54:04  4                   100              100

我认为问题出在"COUNT(DISTINCT Date,IPAddress) AS DistinctDate"作品中。

4

2 回答 2

1

你需要一个COUNT DISTINCT. 这是一个猜测，因为没有提供表结构：

SELECT
  VisitDate,
  COUNT(DISTINCT IPAddress, UniqueID) AS UniqueVisits
FROM MyTable
GROUP BY VisitDate
ORDER BY UniqueVisits DESC

或者，如果您visit date是日期时间或时间戳，请使用DATE函数删除时间部分（注意第二行和第五行的更改）：

SELECT
  DATE(VisitDate),
  COUNT(DISTINCT IPAddress, UniqueID) AS UniqueVisits
FROM MyTable
GROUP BY DATE(VisitDate)
ORDER BY UniqueVisits DESC
于 2013-05-24T20:15:52.277 回答
0

您的日期格式中有时间。所以，我认为所有的日期都是独一无二的。尝试这个：

SELECT
  LicenseID, 
  MAX(Date) AS LatestAccess,
  COUNT(DISTINCT Location) AS DifferentCountries,
  COUNT(DISTINCT IPAddress) AS DistinctIPCount,
  COUNT(DISTINCT date(Date), IPAddress) AS DistinctDate
FROM
  LicenseHistory 
WHERE
  (LicenseID<>30002)
GROUP BY
  LicenseID 
ORDER BY
  DistinctDate DESC
于 2013-05-24T21:54:47.383 回答