Question

我们想要一个SQL语句，列出唯一的IP /唯一ID对在按唯一ID / IP对访问的最大次数排序的任何唯一日期访问过的次数。

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CSV格式表中的一些示例日期：

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

以下是一些示例结果（不正确，最后一列不应该倒数第二）：

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"篇。

Answer 1

您需要COUNT DISTINCT。这是一个猜测，因为没有提供表格结构：

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

或者如果您的visit date是日期时间或时间戳，请使用DATE功能剪掉时间部分（请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

Answer 2

您的日期格式中有一段时间。所以，我认为的所有日期都是唯一的。试试这个：

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

选择不同的ID

2 个答案: