我在表T1中更新了一些计数器值
id, unix_time_stamp, counter1, counter10
1 , 1333435800 , 55 , 80
然后我得到表T2,我在其中复制这些值
id, unix_time_stamp, counter1, counter10, value1, value10
1 , 1333435800 , 55 , 80 , 0 , 0
2 , 1333435801 , 60 , 87 , 5 , 7
3 , 1333435802 , 70 , 90 , 10 , 3
3 , 1333435804 , 80 , 100 , 5 , 5
这是通过一些触发功能完成的
INSERT INTO T2 (unix_time_stamp, counter1, counter10) SELECT unix_time_stamp, counter1, counter10 FROM T1 WHERE id=1
我想要的是将value1,value10计算为
(current_counter1 - last_counter1)/(current_time - last_time)
并将它们放在此插入中
例如带有时间戳1333435804的值1将是
value1=(80-70)/(1333435804-1333435802) = 5
也就是说
insert into t2
(unix_time_stamp, counter1, counter10, value1)
SELECT unix_time_stamp, counter1, counter10,
(counter1 - (select counter1 from T1 order by unix_time_stamp DESC LIMIT 1)/
(unix_time_stamp - (select unix_time_stamp from T1 order by unix_time_stamp DESC LIMIT 1)
FROM T1 WHERE id=1
但是我想要一个短一点的版本,因为我有10个计数器:)
整体情况有点复杂,我有一些理由不在SQL之外执行此操作
我正在使用sqlite
这对我来说太复杂了:)
请帮忙。
最佳答案
我想以下查询将计算您的insert子句所需的所有数据:
SELECT 1e0 * (cur.counter1 - prv.counter1)/(cur.unix_time_stamp - prv.unix_time_stamp) AS [value1]
, 1e0 * (cur.counter10 - prv.counter10)/(cur.unix_time_stamp - prv.unix_time_stamp) AS [value10]
, cur.counter1 AS [cur_counter1], cur.counter10 AS [cur_counter10], cur.unix_time_stamp AS [cur_time]
, prv.counter1 AS [prv_counter1], prv.counter10 AS [prv_counter10], prv.unix_time_stamp AS [prv_time]
FROM T1 cur, T1 prv
WHERE cur.counter1 = (SELECT MAX(aux_0.counter1) FROM T1 aux_0)
AND prv.counter1 = (SELECT MAX(aux_1.counter1) FROM T1 aux_1 WHERE aux_1.counter1 < cur.counter1);
关于sql - 在SQL中使用计数器值计算行之间的差异,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10334147/