Ingres SQL，根据另一列的值查找一列的最大值

Question

我正在使用从其他人继承的脚本处理Ingres数据库。我需要更改脚本以提取最新的start_time和end_time事件的action_times，以及两者之间的差异。在DB的样品下面列出

id_num | version | action_id | action_time 
---------------------------------------------------------------------------- 
1   2   start_time  2014-05-26 14:58:14 
1   2   end_time  2014-05-26 14:58:16 
1   4   start_time  2014-05-27 10:10:57 
1   4   end_time  2014-05-27 10:10:11 

到目前为止，我想出是：

SELECT max(a.action_time) as BIG, max(b.action_time) as SMALL, max(a.action_time) - max(b.action_time) as DIFF 
FROM table1 as a, table1 as b, 
WHERE a.id_num = '1' AND a.action_id = 'end_time' AND b.id_num = '1' AND b.action_id = 'start_time' 

但结果出来如下：

BIG       SMALL     DIFF 
---------------------------------------------------------------------------- 
2014-05-27 10:10:11   2014-05-27 10:10:57  null  

道歉如果这样的问题已经得到解答（我确信它可能有），但我已经花了几天的时间来查看各种论坛，我找不到类似的例子，可能我是如何解释搜索条件的。任何帮助将不胜感激，我敢肯定，我会在大学里涵盖这样的东西，但那是几年前，我的SQL现在有点生疏。提前致谢！

编辑：所以一些研究，我想出了下面这将在DB GUI工作后：如果您想计算最大（a.acction_time）之间的差值

SELECT ingresdate(varchar(max(a.action_time))) as BIG, ingresdate(varchar(max(b.action_time))) as SMALL, date_part('secs',ingresdate(varchar(max(a.action_time))) - ingresdate(varchar(max(b.action_time)))) as DIFF 
FROM table1 as a, table1 as b, 
WHERE a.id_num = '1' AND a.action_id = 'end_time' AND b.id_num = '1' AND b.action_id = 'start_time' 

Answer 1

，和max（ b.acction_time），你应该使用下面的脚本：

SELECT max(a.acction_time) as BIG, max(b.acction_time) as SMALL,DATEDIFF(s, max(a.acction_time), max(b.acction_time)) as DIFF 
FROM table1 as a, table1 as b 
WHERE a.id_num = '1' AND a.action_id = 'end_time' AND b.id_num = '1' AND b.action_id = 'start_time' 

如果你不记得DATEDIFF（）函数，我会解释给你。

P.S：你的table1中的主键在哪里？

Answer 2

我会为此使用子选择。尝试： -

select a.action_time as max_end_time, b.action_time as max_start_time, 
a.action_time - b.action_time as diff 
from table a, table b 

where a.action_time = (select max(action_time) 
from table where action_id = 'end_time') 

and b.action_time = (select max(action_time) 
from table where action_id = 'start_time) 

Answer 3

这里是我的尝试：

SELECT start.action_time, end.action_time, 
    interval('seconds', end.action_time - start.action_time) as diff_secs 
FROM 
(
SELECT action_time 
FROM table a 
INNER JOIN 
( SELECT max(id_num) as max_id_num, max(version) as max_version FROM table 
) b on (id_num = max_id_num and version = max_version) 
WHERE a.action_id = 'start_time' 
) start 
CROSS JOIN 
(
SELECT action_time 
FROM table a 
INNER JOIN 
( SELECT max(id_num) as max_id_num, max(version) as max_version FROM table 
) b on (id_num = max_id_num and version = max_version) 
WHERE a.action_id = 'end_time' 
) end 

使用您的数据得到以下的输出：

+----------------------+----------------------+-----------+ 
|  action_time  |  action_time  | diff_secs | 
+----------------------+----------------------+-----------+ 
| 27-May-2014 10:10:57 | 27-May-2014 10:10:11 |  -46 | 
+----------------------+----------------------+-----------+ 

仅供参考，这里是我用来创建脚本，填充测试表

CREATE TABLE table 
(
id_num integer, 
version integer, 
action_id char(10), 
action_time timestamp 
) 

INSERT INTO table VALUES (1,2,'start_time', '2014-05-26 14:58:14'); 
INSERT INTO table VALUES (1,2,'end_time', '2014-05-26 14:58:16'); 
INSERT INTO table VALUES (1,4,'start_time', '2014-05-27 10:10:57'); 
INSERT INTO table VALUES (1,4,'end_time', '2014-05-27 10:10:11');