从mysql表中获得第二高的值

Question

我有一个包含四个字段的表,如下所示,

(UID是用户ID)

ID UID MUSIC DATE 
1   0   a   2013-10-20
2   0   a   2013-10-21
3   0   a   2013-10-22
4   0   a   2013-10-24
5   0   b   2013-10-11
8   0   b   2013-10-15
10  0   c   2013-10-26
9   0   c   2013-10-25
7   0   c   2013-10-20
6   0   c   2013-10-18
11  0   d   2013-10-10

如何使用MySQL Query从上表中检索所有第二高的日期？

预期结果:

ID UID MUSIC DATE 
3   0   a   2013-10-22
5   0   b   2013-10-11
9   0   c   2013-10-25

要么

ID UID MUSIC DATE 
3   0   a   2013-10-22
5   0   b   2013-10-11
9   0   c   2013-10-25
11  0   d   2013-10-10

Answer 1

好的,我想我有答案.请检查一下:

SELECT tbl.ID, tbl.UID, tbl.MUSIC, tbl.DATE
FROM tbl
INNER JOIN
(
SELECT MUSIC,
       Substring_index(Substring_index(gdate, ',', 2), ',', -1) AS sec_date
FROM   (SELECT MUSIC               ,
               GROUP_CONCAT(DATE order by DATE desc separator ",") AS gdate
        FROM   tbl
        GROUP  BY MUSIC) t1
) AS tbl2
  ON tbl.MUSIC=tbl2.MUSIC
    AND tbl.DATE=tbl2.sec_date

首先,我在DATE创建了GROUP_CONCAT,由desc命令,所以我可以使用它Substring_index来获取第二个DATE,当然,按MUSIC对所有内容进行分组,因此日期按照各自的MUSIC类别进行分组.然后我写了实际的查询以获得结果,并连接到派生表,所以我确保我得到该特定MUSICAND的正确行DATE.

这是SQLFiddle

UPDATE

如果您想通过UID进一步过滤,只需添加WHERE到内部查询,如下所示:

SELECT tbl.ID, tbl.UID, tbl.MUSIC, tbl.DATE
FROM tbl
INNER JOIN
(
SELECT MUSIC,
       Substring_index(Substring_index(gdate, ',', 2), ',', -1) AS sec_date
FROM   (SELECT MUSIC               ,
               GROUP_CONCAT(DATE order by DATE desc separator ",") AS gdate
        FROM   tbl
        WHERE UID=1 -- add filter here
        GROUP  BY MUSIC) t1
) AS tbl2
  ON tbl.MUSIC=tbl2.MUSIC
    AND tbl.DATE=tbl2.sec_date

并更新了SQLFiddle