SQL查询前值的每一行

Question

我有以下结构的SQL表：

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

所以一些样本行是：

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

我需要生成以下格式输出：

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

所以我输出的格式是：

id Integer, 
version Integer, 
attribute_current varchar, 
attribute_old varchar 

我已经试过如下：

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); 

上面的查询执行，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。

约我应该怎么解决我的查询产生正确的结果任何想法？谢谢！

注 - 版本号，但不保证顺序从1开始的我实际的数据库中有一些不寻常的版本号（即用户7有3版和第15版没有4，5，6，等...） 。

Answer 1

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; 

样本输出（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 2

既然你已经提到：

版本号并不能保证在1我实际的数据库是连续的起点有一些不寻常的版本号（即用户7有3版和第15版无4，5，6，等...）

MySQL不支持像任何其他RDBMS窗口功能呢，你仍然可以模拟你如何创建一个序列号，并作为连接列获取以前的行。当然，

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 3

如果版本号总是1增加，你可以：

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; 

提供的版本值是按数字顺序排列。

Answer 5

我想表达这种最简单的方法是用相关子查询：

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 

该版本将放在NULL为第一行的上一个值。如果你真的想重复：

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