(有关本文的工作示例,请参阅SQLFiddle )
我有一个包含股票信息的表,如下所示:
sp100_id _date bullishness agreement
----------------------------------------------
1 2011-03-16 1.01 0.33
1 2011-03-17 0.85 1.28
1 2011-03-18 0.89 1.25
1 2011-03-21 1.46 1.21
1 2011-03-22 0.39 2.53
2 2011-03-16 3.07 1.27
2 2011-03-17 2.09 0.80
2 2011-03-18 0.91 0.12
2 2011-03-21 1.50 0.00
2 2011-03-22 2.62 1.10
3 2011-03-16 0.73 1.13
3 2011-03-17 1.13 1.21
3 2011-03-18 1.12 0.45
3 2011-03-21 1.00 1.01
3 2011-03-22 1.00 0.53
4 2011-03-16 0.40 1.10
4 2011-03-17 2.40 0.03
4 2011-03-18 3.16 0.10
4 2011-03-21 0.86 0.50
4 2011-03-22 1.00 0.10我需要将这些公司(sp100_id)按其随着时间推移的平均bullishness排在前3位:
SELECT
sp100_id,
AVG(bullishness) as bullishness,
AVG(agreement) AS agreement
FROM stocks
WHERE _date BETWEEN '2011-03-16' AND '2011-03-22'
GROUP BY sp100_id LIMIT 3这可以很好地工作,因为结果是
SP100_ID BULLISHNESS AGREEMENT
2 2.038 0.658
4 1.564 0.366
3 0.996 0.866现在我有了前3名,我需要按AGREEMENT升序对前3名进行重新排序:
SP100_ID BULLISHNESS AGREEMENT
4 1.564 0.366
2 2.038 0.658
3 0.996 0.866一个查询就可以做到这一点吗?我尝试了以下方法,但不起作用。它仍然只按bullishness订购
SELECT
sp100_id,
AVG(bullishness) as bullishness,
AVG(agreement) AS agreement
FROM stocks
WHERE _date BETWEEN '2011-03-16' AND '2011-03-22'
GROUP BY sp100_id
ORDER BY bullishness DESC, agreement ASC LIMIT 3因此,需要明确的是:(1)我需要找到平均看涨程度最高的前3家公司(2)这3家公司需要从最低到最高的一致性排序。优选地,使用一个查询。你知道怎么做吗?
发布于 2012-09-02 19:53:07
它被称为结构化查询语言,因为您可以构建将查询(也称为虚拟表)嵌套在其他查询中的结构。
第一个查询是正确的,但它需要自己的ORDER BY子句,并将其嵌套在另一个查询中,如下所示。
SELECT *
FROM (
SELECT sp100_id,
AVG(bullishness) as bullishness,
AVG(agreement) AS agreement
FROM stocks
WHERE _date BETWEEN '2011-03-16' AND '2011-03-22'
GROUP BY sp100_id
ORDER BY bullishness DESC
LIMIT 3
) subquery
ORDER BY agreement ASC去拉小提琴:http://sqlfiddle.com/#!2/c9ff0/7/0
https://stackoverflow.com/questions/12235509
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