Question

我有一个具有以下结构的SQL表：

id Integer, (represents a userId)
version Integer,
attribute varchar(50)

所以一些示例行是：

4, 1, 'test1'
4, 2, 'test2'
4, 3, 'test3'

我需要按以下格式生成输出：

4, 1, 'test1', 'test1'
4, 2, 'test2', 'test1'
4, 3, 'test3', 'test2'

所以输出的格式是：

id Integer,
version Integer,
attribute_current varchar,
attribute_old varchar

我已经尝试了以下内容：

select versionTest.id, versionTest.version, versionTest.attribute, maxVersionTest.attribute
from versionTest
inner join
versionTest maxVersionTest
ON
versionTest.id = versionTest.id
AND
versionTest.version = 
(Select max(version) currentMaxVersion 
from versionTest maxRow
where maxRow.id = id);

执行上述查询，但返回不正确的结果。它只返回最大版本，而不是返回所有版本的行。

有关如何修复查询以产生正确结果的任何想法？谢谢！

注意 - 版本号不保证从1开始是连续的。我的实际数据库有一些不寻常的版本号（即用户7有版本3和15，没有版本4,5,6等...）。

Answer 1

既然你提到过：

版本号不保证从1开始是连续的。我的实际数据库有一些不寻常的版本号（即用户7有版本3和15，没有版本4,5,6等...）

MySQL不像任何其他RDBMS那样支持窗口函数，你仍然可以模拟如何创建序列号并用作连接列来获取前面的行。前，

SELECT  a.ID, a.Version, a.attribute attribute_new,
        COALESCE(b.attribute, a.attribute) attribute_old
FROM
        (
            SELECT  ID, version, attribute,
                    @r1 := @r1 + 1 rn
            FROM    TableName, (SELECT @r1 := 0) b
            WHERE   ID = 4
            ORDER   BY version
        ) a
        LEFT JOIN
        (
            SELECT  ID, version, attribute,
                    @r2 := @r2 + 1 rn
            FROM    TableName, (SELECT @r2 := 0) b
            WHERE   ID = 4
            ORDER   BY version
        ) b ON a.rn = b.rn + 1

SQLFiddle Demo

Answer 2

SELECT a.*, COALESCE(b.attribute,a.attribute) attribute_old
  FROM
     ( SELECT x.*
            , COUNT(*) rank 
         FROM versiontest x 
         JOIN versiontest y 
           ON y.id = x.id 
          AND y.version <= x.version 
        GROUP 
           BY x.id
            , x.version
     ) a
  LEFT
  JOIN
     ( SELECT x.*
            , COUNT(*) rank 
         FROM versiontest x 
         JOIN versiontest y 
           ON y.id = x.id 
          AND y.version <= x.version 
        GROUP 
           BY x.id
            , x.version
     ) b
    ON b.id = a.id 
   AND b.rank = a.rank-1;

示例输出（DEMO）：

+----+---------+-----------+------+---------------+
| id | version | attribute | rank | attribute_old |
+----+---------+-----------+------+---------------+
|  4 |       1 | test1     |    1 | test1         |
|  4 |       5 | test2     |    2 | test1         |
|  4 |       7 | test3     |    3 | test2         |
|  5 |       2 | test3     |    1 | test3         |
|  5 |       3 | test4     |    2 | test3         |
|  5 |       8 | test5     |    3 | test4         |
+----+---------+-----------+------+---------------+

Answer 3

如果版本号总是增加1，您可以：

select  cur.id
,       cur.version
,       cur.attribute
,       coalesce(prev.attribute, cur.attribute)
from    versionTest
left join
        versionTest prev
on      prev.id = cur.id
        and prev.version = cur.version + 1

Answer 4

你可以试试......

SELECT ID,
       VERSION,
       ATTRIBUTE,
       (SELECT ATTRIBUTE
            FROM VERSIONTEST V3
            WHERE V3.ID = V1.ID AND
                  V3.VERSION = (SELECT MAX(VERSION)
                                    FROM VERSIONTEST V2
                                    WHERE V2.ID = V1.ID AND
                                          V2.VERSION < V1.VERSION)) AS PREVIOUSATTRIBUTE
    FROM VERSIONTEST V1;

如果版本值按数字顺序排列。

Answer 5

我认为最简单的表达方式是使用相关的子查询：

select id, version, attribute as attribute_current,
       (select attribute
        from VersionTest vt2
        where vt2.id = vt.id and vt2.version < vt.version
        order by vt2.version
        limit 1
       ) as attribute_prev
from VersionTest vt

此版本将NULL作为第一行的prev值。如果你真的想重复它：

select id, version, attribute as attribute_current,
       coalesce((select attribute
                 from VersionTest vt2
                 where vt2.id = vt.id and vt2.version < vt.version
                 order by vt2.version
                 limit 1
                ), vt.attribute
               ) as attribute_prev
from VersionTest vt

SQL查找每行的上一个值

5 个答案: