从由其他列聚合并选择最大日期的列中获取任何值

Question

我是 SQL 新手，我有一个包含以下列的表

+-------+--------+----+----+
| email |  date  | IP | ID |
+-------+--------+----+----+

我想做这样的事情：

SELECT T.email,
       Max(T.date),
       T.ip     AS User_IP,
       T.id     AS ID

FROM LoginTable as T

WHERE (IP IS NOT NULL)

GROUP BY T.email

这当然行不通，因为 IP 和 ID 不在（最大）聚合函数中。但我需要 IP 与 ID 和日期匹配，所以我不能在这些列中使用 (max)，因为我会从不同的行中得到结果，这不是一个选项。

回顾一下，我需要：

• 每封唯一电子邮件有一行（因此是分组依据）
• 所选行是日期最近的行 -> max(date)
• 我需要 IP 和 ID 与 max(date) 选择的行位于同一行。

Answer 1

救援窗口函数：

SELECT email, date, user_ip, id
FROM   (SELECT email, date, ip AS user_ip, logid AS id, 
               ROW_NUMBER() OVER (PARTITION BY email ORDER BY date DESC) AS rn
        FROM   mytable
        WHERE  ip IS NOT NULL) t
WHERE  rn = 1

注意：此查询将准确检索每封唯一电子邮件的一行。如果你想支持与最大日期绑定的记录，你应该使用rank() 而不是row_number()。

Answer 2

您可以使用 sql server 中可用的Ranking Functions 代替GROUP BY。

一个例子是ROW_NUMBER()

;WITH CTE AS (
 SELECT  ROW_NUMBER () OVER (PARTITION BY email ORDER BY DATE DESC) AS rn,
         email, 
         date,
         ip        AS User_IP,
         Logid     AS ID        
 FROM Table T WHERE (IP IS NOT NULL)
)
SELECT * FROM CTE WHERE RN=1

Answer 3

如果您在 Table(email, date) 上有索引，这将非常有用

SELECT email  AS email,
  date        AS Date,
  ip          AS User_IP,
  Logid       AS ID
FROM Table
WHERE IP IS NOT NULL and 
      (email, date) = (
           SELECT email, Max(date) 
           FROM Table 
           GROUP BY email
       )

Answer 4

试试这个答案，

SELECT L.email  AS email,
L.date      AS Date,
L.ip        AS User_IP,
L.Logid     AS ID
FROM Table AS T
WHERE EXISTS(
    SELECT email,MAX(Date)Date 
    FROM Table 
    GROUP BY email) AND (IP IS NOT NULL)
GROUP BY T.email

希望对你有所帮助。