Question

我在这里面临一个角落......

背景：

TABLE myrecord (
 id     int         # primary key
 name   varchar(32) # test name
 d_when varchar(8)  # date in yyyymmdd string format
)

内容：

id    name     d_when
100   Alan     20110201
101   Dave     20110304
102   Alan     20121123
103   Alan     20131001
104   Dave     20131002
105   Bob      20131004
106   Mike     20131101

在外行人看来，我想知道谁是“回归者”，什么时候是他的最后一次（即“倒数第二次”）访问。像热情的东西：

SELECT SECOND_MAX(id), CORRESPONDING(d_when) 
FROM myrecord 
GROUP BY name 
HAVING count(name)>1;

结果预期：

101   Dave     20110304 
102   Alan     20121123

到目前为止，我尝试了以下内容。

SELECT T1.id, t1.name, T1.d_when 
FROM myrecord t1 
WHERE id IN (SELECT MAX(id), 
    COUNT(id) cn 
    WHERE cn>1 
    ORDER BY d_when DESC)

但有些事情显然不在这里。

Answer 1

这是一种方式......

SELECT x.* 
  FROM my_table x 
  JOIN my_table y 
    ON y.name=x.name 
   AND y.d_when >= x.d_when 
 GROUP 
    BY x.name 
     , x.d_when 
HAVING COUNT(*) = 2;

Answer 2

为什么第二个最后一个id是必要的？

如果不是：

SELECT MAX(id), name, MAX(d_when)
FROM myrecord
    GROUP BY name 
    HAVING count(name)>1

如果是：

SELECT name, max(id), max(d_when)
FROM myrecord
WHERE 
-- get only the names that have more then one occurrence
name in (
    SELECT name
    FROM myrecord
    GROUP BY name
    HAVING COUNT(*) > 1
)
-- filter out the max id's
AND id NOT IN(
    SELECT max(id)
    FROM myrecord
    GROUP BY name
)
GROUP BY name

甚至更好（感谢@Andomar的提及）：

SELECT name, max(id), max(d_when)
FROM myrecord
WHERE 
-- filter out the max id's, this will also filter out those with one record
AND id NOT IN(
    SELECT max(id)
    FROM myrecord
    GROUP BY name
)
GROUP BY name

Answer 3

在MySQL中，用于检索所有第二次访问和第二次访问日期的人。

查询：

SELECT *
FROM
(
SELECT
   @ID:=CASE WHEN @Name <> Name THEN 1 ELSE @ID+1 END AS rn,
   @Name:=Name AS Name,
   ID, d_when
FROM
  (SELECT ID, Name, d_when
   FROM myrecord
   ORDER BY Name asc, d_when asc
  ) rec1,   (SELECT @ID:= 0) rec1_id, (SELECT @Name:= 0) rec1_nm
) rec
where rec.rn=2

输出：

rn  Name    ID  d_when
2   Dave    104 20131002
2   Alan    102 20121123

Answer 4

假设id列随着时间的推移而上升，您可以为每个名称选择次高d_when，如：

select  name
,       d_when
from    YourTable yt1
where   id in
        (
        select  max(id)
        from    YourTable yt2
        where   id not in
                (
                select  max(id)
                from    YourTable yt3
                group by
                        name
                )
        group by
                name
        )

Answer 5

select * from 
myrecord
where id in (
    SELECT max(id)
    FROM myrecord 
    WHERE id not in (SELECT MAX(id)
        FROM myrecord   
        GROUP BY name 
        HAVING count(name)>1)
    group by name )

在按col1分组时获取col2的第二个最大值

5 个答案: