就像已经写在标题中一样，我首先将每个客户的所有销售额加上去年和今年，然后从去年开始减去{{1} }为每个客户。
修改：
(last year - this year)
所以去年安娜的成绩应该是400，而今年600。差异应该是200。
我已经尝试了一些查询但对我来说太复杂了：
clientid | ....date..... | quant | product. | client | place | price |
......1... |2016-01-10 | ...1.... | product1 | anna | Italy... | .100. |
......1... |2016-04-12 | ...2.... | product2 | anna | Italy... | .300. |
......1... |2017-01-10 | ...1.... | product1 | anna | Italy... | .200. |
......1... |2017-04-12 | ...2.... | product1 | anna | Italy... | .400. |
答案 0 :(得分：0)
可能你可以试试这个。在一个子查询中，我将当前年份的总和和另一个子查询取上一年的总和，并使用两者来得到不同的。
Select
m1.clientid,
m1.cyear,
m0.pyear,
m1.sum csum,
m0.sum psum,
(
m1.sum - m0.sum
)
diff
from
(
select
clientid,
cyear,
sum
from
(
select
clientid,
year(date) cyear,
sum(price) sum
from
buget
group by
clientid,
cyear
)
T
where
cyear = year(curdate())
)
m1,
(
(
select
clientid, pyear, sum
from
(
select
clientid,
year(date) pyear,
sum(price) sum
from
buget
group by
clientid,
pyear
)
T
where
pyear =
(
year(curdate())
)
- 1) m0
)
where
m1.clientid = m0.clientid
答案 1 :(得分：0)
更新了答案
select ABS(t3.price2 - t2.price1),t1.client from buget t1 inner join (select sum(price) as price1 from buget where year(date) = year(now()) group by client) t2 on year(date) = year(now()) inner join (select sum(price) as price2 from buget where year(date) = year(now()) - 1 group by client) t3 group by client;