Question

我想使用下表中相应的REVID为每个NID检索最后一个“VERSION”：

+------+-------+-----+---------+----------+
| RHID | REVID | NID | VERSION |  ACTION  |
+------+-------+-----+---------+----------+ 
| 1    |   1   |  1  |    1    |  CREATE  |
+------+-------+-----+---------+----------+  
| 2    |   2   |  2  |    1    |  CREATE  |
+------+-------+-----+---------+----------+  
| 3    |   3   |  1  |    2    |  UPDATE  |
+------+-------+-----+---------+----------+  
| 4    |   4   |  1  |    3    |  UPDATE  |
+------+-------+-----+---------+----------+  
| 16   |   3   |  1  |    4    | ROLLBACK |
+------+-------+-----+---------+----------+

当我跑步时：

SELECT a.nid, a.revID, MAX(a.version)
FROM `revision_history` a
GROUP BY a.nid;

我希望结果是：

+-----+-------+---------+
| NID | REVID | VERSION |
+-----+-------+---------+
|  1  |   3   |    4    |
+-----+-------+---------+
|  2  |   2   |    1    |
+-----+-------+---------+

相反，结果是：

+-----+-------+---------+
| NID | REVID | VERSION |
+-----+-------+---------+
|  1  |   3   |    4    |
+-----+-------+---------+
|  2  |   2   |    4    |
+-----+-------+---------+

问。获得我期望的结果的适当查询是什么？

注：

这是为了构建一个视图（在MySQL中，它不支持子查询）。

Answer 1

试试这个：

select a.nid ,a.revID ,a.version
from   `revision_history` a
JOIN
      (SELECT nid,  MAX(version) as version
        FROM `revision_history` 
        GROUP BY nid)b
ON    a.nid=b.nid
AND   a.version=b.version

SQL fiddle demo

Answer 2

我总是倾向于认为GROUP BY取FROM中的第一个值：

SELECT nid ,revID ,version
FROM (  SELECT nid, revID, version
        FROM `revision_history`
        ORDER BY version DESC) AS h
GROUP BY nid

如果您了解行为并且永远不会超过1个案例，则该方法非常简短，易于理解和编辑。如果由于某种原因，您在同一nid上会有2个完全相同的版本，那么结果可能是错误的。

使用MAX（列）时，其他列的值不匹配

2 个答案:

SQL fiddle demo