Question

编辑：

我的表格由id，segmentID，月份，年份和平均速度组成，我在这里创建了一些带有一些演示数据的sqlfiddle：http://sqlfiddle.com/#!2/183c1/1

我需要确定平均驾驶速度在过去一个月内下降超过10％的细分市场。

有人可以帮我吗？感谢

由于

Answer 1

您需要选择每个月，加入下个月（由于您的表结构而有点复杂）并找到减少（/增加）。请尝试以下复杂查询

SELECT 
   t1.segmentID, t1.month, t1.year, AVG(t1.avgSpeed) as avgSpeed1, 
   AVG(t2.avgSpeed) as avgSpeed2,
   1-(AVG(t1.avgSpeed)/AVG(t2.avgSpeed)) as decrease
FROM 
   travels t1
LEFT JOIN 
   travels t2
ON 
   CONCAT(t2.year,'-',LPAD(t2.month,2,'00'),'-',LPAD(1,2,'00')) = DATE_ADD(CONCAT(t1.year,'-',LPAD(t1.month,2,'00'),'-',LPAD(1,2,'00')), INTERVAL -1 MONTH)
GROUP BY 
   segmentID, month, year
HAVING 
   avgSpeed1/avgSpeed2 < .9

这是更新后的SQLFiddle - http://sqlfiddle.com/#!2/183c1/25

Answer 2

这是我的解决方案

Sqlfidle demo

关键是要记录上个月和下一个月之间，所以我做了一年* 100 +月和一年又一年，并且检查差异1和89年* 100 +月场。另外，MySQL不支持CTE并使用派生表使查询变得丑陋也很可怜。

代码：

select s.month,s.speed,m.month as prevmonth,m.speed as sp, 100-s.speed/m.speed*100 as speeddiff  from
(SELECT segmentid,month,year*100+month as mark,avg(avgSpeed) as speed from travels
group by segmentid,month,year*100+month
) as s
,
(SELECT segmentid,month,year*100+month as mark,avg(avgSpeed) as speed from travels
group by segmentid,month,year*100+month
) as m   
where s.segmentid=m.segmentid and (s.mark=m.mark+1 or s.mark=m.mark+89) and (m.speed-(m.speed/10))>s.speed;

CTE代码处理除MySQL以外的每个数据库

with t  as(SELECT segmentid,month,year*100+month as mark,avg(avgSpeed) as speed from travels
group by segmentid,month,year*100+month
)
select s.month,s.speed,m.month as prevmonth,m.speed as sp, 100-s.speed/m.speed*100 as speeddiff from t s 
inner join t m on s.segmentid=m.segmentid and (s.mark=m.mark+1 or s.mark=m.mark+89)
where (m.speed-(m.speed/10))>s.speed;

Answer 3

这需要自我加入。这个答案将帮助您入门。你可以处理细节。

select somefields
from yourtable t1 join yourtable t2 on t1.something = t2.something
where t1.month = whatever
and t2.month = t1.month + 1
and t2.speed <= t1.speed * .9

SQL复杂的查询

3 个答案: