Question

我有销售表并有两个不同的日期范围。即，我的总销售额在(2016-12-21 - 2016-12-30) is 100和期间(2016-12-11 - 2016-12-20) is 85之间。现在我想要的结果是

100 (sales of 2016-12-21 - 2016-12-30), 85 (sales of 2016-12-11 - 2016-12-20), 15 (difference of both periods)通过单一查询。

我在想什么

select *, (a.sales - b.sales) as diff 
from (select id, sum(sales) as sales from salestable where date >= '2016-12-21' and date <= '2016-12-30') a 
join (select id, sum(sales) as sales from salestable where date >= '2016-12-11' and date <= '2016-12-20') b
on a.id = b.id;

还有其他更好的方法吗？

Answer 1

您可以使用条件聚合：

select sum(case when date >= '2016-12-21' and date <= '2016-12-30' then sales else 0
           end) as sales_a,
       sum(case when date >= '2016-12-11' and date <= '2016-12-20' then sales else 0
           end) as sales_b,
       sum(case when date >= '2016-12-21' and date <= '2016-12-30'
                then sales else 0
                when date >= '2016-12-11' and date <= '2016-12-20'
                then -sales
                else 0
           end) as sales_diff
from salestable;

如果您想要id的总和（根据您id的建议），请将id添加到select并添加group by id

Answer 2

您可以使用case来执行这样的条件求和：

select id,
    sum_21_to_30, 
    sum_11_to_20, 
    sum_21_to_30 - sum_11_to_20 diff
from (select id,
    sum(case when date >= '2016-12-21' and date <= '2016-12-30' then sales else 0 end) sum_21_to_30,
    sum(case when date >= '2016-12-11' and date <= '2016-12-20' then sales else 0 end) sum_11_to_20
from table group by id) t;

在单个查询中获取2个不同日期范围的差异记录

2 个答案: