我已经从几个来源获得了一份温度样本表，我希望在设定的时间间隔内找到所有来源的最小，最大和平均温度。乍一看，这很容易就像这样：
SELECT MIN(temp), MAX(temp), AVG(temp) FROM samples GROUP BY time;
然而，事情变得更加复杂（到了我难以接受的地步！）如果来源进出，而不是在有问题的时间间隔内忽略缺失的来源我想使用来源的最后知道的温度对于遗失的样本。使用日期时间和在不均匀分布的样本中构建间隔（比如说每分钟）会使事情变得复杂化。
我认为应该可以通过在样本表上进行自联接来创建我想要的结果，其中第一个表的时间大于或等于第二个表的时间，然后计算聚合值按源分组的行。但是，我对如何实际做到这一点感到难过。
这是我的测试表：
+------+------+------+
| time | source | temp |
+------+------+------+
| 1 | a | 20 |
| 1 | b | 18 |
| 1 | c | 23 |
| 2 | b | 21 |
| 2 | c | 20 |
| 2 | a | 18 |
| 3 | a | 16 |
| 3 | c | 13 |
| 4 | c | 15 |
| 4 | a | 4 |
| 4 | b | 31 |
| 5 | b | 10 |
| 5 | c | 16 |
| 5 | a | 22 |
| 6 | a | 18 |
| 6 | b | 17 |
| 7 | a | 20 |
| 7 | b | 19 |
+------+------+------+
INSERT INTO samples (time, source, temp) VALUES (1, 'a', 20), (1, 'b', 18), (1, 'c', 23), (2, 'b', 21), (2, 'c', 20), (2, 'a', 18), (3, 'a', 16), (3, 'c', 13), (4, 'c', 15), (4, 'a', 4), (4, 'b', 31), (5, 'b', 10), (5, 'c', 16), (5, 'a', 22), (6, 'a', 18), (6, 'b', 17), (7, 'a', 20), (7, 'b', 19);
要进行min，max和avg计算，我想要一个如下所示的中间表：
+------+------+------+
| time | source | temp |
+------+------+------+
| 1 | a | 20 |
| 1 | b | 18 |
| 1 | c | 23 |
| 2 | b | 21 |
| 2 | c | 20 |
| 2 | a | 18 |
| 3 | a | 16 |
| 3 | b | 21 |
| 3 | c | 13 |
| 4 | c | 15 |
| 4 | a | 4 |
| 4 | b | 31 |
| 5 | b | 10 |
| 5 | c | 16 |
| 5 | a | 22 |
| 6 | a | 18 |
| 6 | b | 17 |
| 6 | c | 16 |
| 7 | a | 20 |
| 7 | b | 19 |
| 7 | c | 16 |
+------+------+------+
以下查询让我接近我想要的但它需要源的第一个结果的温度值，而不是给定时间间隔的最新结果：
SELECT s.dt as sdt, s.mac, ss.temp, MAX(ss.dt) as maxdt FROM (SELECT DISTINCT dt FROM samples) AS s CROSS JOIN samples AS ss WHERE s.dt >= ss.dt GROUP BY sdt, mac HAVING maxdt <= s.dt ORDER BY sdt ASC, maxdt ASC;
+------+------+------+-------+
| sdt | mac | temp | maxdt |
+------+------+------+-------+
| 1 | a | 20 | 1 |
| 1 | c | 23 | 1 |
| 1 | b | 18 | 1 |
| 2 | a | 20 | 2 |
| 2 | c | 23 | 2 |
| 2 | b | 18 | 2 |
| 3 | b | 18 | 2 |
| 3 | a | 20 | 3 |
| 3 | c | 23 | 3 |
| 4 | a | 20 | 4 |
| 4 | c | 23 | 4 |
| 4 | b | 18 | 4 |
| 5 | a | 20 | 5 |
| 5 | c | 23 | 5 |
| 5 | b | 18 | 5 |
| 6 | c | 23 | 5 |
| 6 | a | 20 | 6 |
| 6 | b | 18 | 6 |
| 7 | c | 23 | 5 |
| 7 | b | 18 | 7 |
| 7 | a | 20 | 7 |
+------+------+------+-------+
更新： chadhoc（伟大的名字，顺便说一下！）提供了一个很好的解决方案，遗憾的是在MySQL中不起作用，因为它不支持他使用的FULL JOIN。幸运的是，我相信一个简单的UNION是一个有效的替代品：
-- Unify the original samples with the missing values that we've calculated
(
SELECT time, source, temp
FROM samples
)
UNION
( -- Pull all the time/source combinations that we are missing from the sample set, along with the temp
-- from the last sampled interval for the same time/source combination if we do not have one
SELECT a.time, a.source, (SELECT t2.temp FROM samples AS t2 WHERE t2.time < a.time AND t2.source = a.source ORDER BY t2.time DESC LIMIT 1) AS temp
FROM
( -- All values we want to get should be a cross of time/temp
SELECT t1.time, s1.source
FROM
(SELECT DISTINCT time FROM samples) AS t1
CROSS JOIN
(SELECT DISTINCT source FROM samples) AS s1
) AS a
LEFT JOIN samples s
ON a.time = s.time
AND a.source = s.source
WHERE s.source IS NULL
)
ORDER BY time, source;
更新2： MySQL为chadhoc的代码提供以下EXPLAIN输出：
+----+--------------------+------------+------+---------------+------+---------+------+------+-----------------------------+
| id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
+----+--------------------+------------+------+---------------+------+---------+------+------+-----------------------------+
| 1 | PRIMARY | temp | ALL | NULL | NULL | NULL | NULL | 18 | |
| 2 | UNION | <derived4> | ALL | NULL | NULL | NULL | NULL | 21 | |
| 2 | UNION | s | ALL | NULL | NULL | NULL | NULL | 18 | Using where |
| 4 | DERIVED | <derived6> | ALL | NULL | NULL | NULL | NULL | 3 | |
| 4 | DERIVED | <derived5> | ALL | NULL | NULL | NULL | NULL | 7 | |
| 6 | DERIVED | temp | ALL | NULL | NULL | NULL | NULL | 18 | Using temporary |
| 5 | DERIVED | temp | ALL | NULL | NULL | NULL | NULL | 18 | Using temporary |
| 3 | DEPENDENT SUBQUERY | t2 | ALL | NULL | NULL | NULL | NULL | 18 | Using where; Using filesort |
| NULL | UNION RESULT | <union1,2> | ALL | NULL | NULL | NULL | NULL | NULL | Using filesort |
+----+--------------------+------------+------+---------------+------+---------+------+------+-----------------------------+
我能够让Charles的代码像这样工作：
SELECT T.time, S.source,
COALESCE(
D.temp,
(
SELECT temp FROM samples
WHERE source = S.source AND time = (
SELECT MAX(time)
FROM samples
WHERE
source = S.source
AND time < T.time
)
)
) AS temp
FROM (SELECT DISTINCT time FROM samples) AS T
CROSS JOIN (SELECT DISTINCT source FROM samples) AS S
LEFT JOIN samples AS D
ON D.source = S.source AND D.time = T.time
其解释是：
+----+--------------------+------------+------+---------------+------+---------+------+------+-----------------+
| id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
+----+--------------------+------------+------+---------------+------+---------+------+------+-----------------+
| 1 | PRIMARY | <derived5> | ALL | NULL | NULL | NULL | NULL | 3 | |
| 1 | PRIMARY | <derived4> | ALL | NULL | NULL | NULL | NULL | 7 | |
| 1 | PRIMARY | D | ALL | NULL | NULL | NULL | NULL | 18 | |
| 5 | DERIVED | temp | ALL | NULL | NULL | NULL | NULL | 18 | Using temporary |
| 4 | DERIVED | temp | ALL | NULL | NULL | NULL | NULL | 18 | Using temporary |
| 2 | DEPENDENT SUBQUERY | temp | ALL | NULL | NULL | NULL | NULL | 18 | Using where |
| 3 | DEPENDENT SUBQUERY | temp | ALL | NULL | NULL | NULL | NULL | 18 | Using where |
+----+--------------------+------------+------+---------------+------+---------+------+------+-----------------+
答案 0 :(得分：1)
我认为使用mySql中的排名/窗口函数可以获得更好的性能，但不幸的是我不知道这些以及TSQL实现。以下是符合ANSI标准的解决方案：
-- Full join across the sample set and anything missing from the sample set, pulling the missing temp first if we do not have one
select coalesce(c1.[time], c2.[time]) as dt, coalesce(c1.source, c2.source) as source, coalesce(c2.temp, c1.temp) as temp
from samples c1
full join ( -- Pull all the time/source combinations that we are missing from the sample set, along with the temp
-- from the last sampled interval for the same time/source combination if we do not have one
select a.time, a.source,
(select top 1 t2.temp from samples t2 where t2.time < a.time and t2.source = a.source order by t2.time desc) as temp
from
( -- All values we want to get should be a cross of time/samples
select t1.[time], s1.source
from
(select distinct [time] from samples) as t1
cross join
(select distinct source from samples) as s1
) a
left join samples s
on a.[time] = s.time
and a.source = s.source
where s.source is null
) c2
on c1.time = c2.time
and c1.source = c2.source
order by dt, source
答案 1 :(得分：0)
我知道这看起来很复杂，但它的格式是解释自己...... 它应该工作...希望你只有三个来源...如果你有任意数量的来源，这将无法工作......在这种情况下，请看第二个查询...... 编辑：删除了第一次尝试
编辑：如果您提前不知道来源，则必须执行某些操作，创建一个“填写”缺失值的中间结果集。 像这样的东西：
第二次编辑：通过移动逻辑从Select子句中检索每个源的最新临时读数到加入条件，删除了对Coalesce的需求。
Select T.Time, Max(Temp) MaxTemp,
Min(Temp) MinTemp, Avg(Temp) AvgTemp
From
(Select T.TIme, S.Source, D.Temp
From (Select Distinct Time From Samples) T
Cross Join
(Select Distinct Source From Samples) S
Left Join Samples D
On D.Source = S.Source
And D.Time =
(Select Max(Time)
From Samples
Where Source = S.Source
And Time <= T.Time)) Z
Group By T.Time