SQL如何查询总计和小计
Posted
技术标签:
【中文标题】SQL如何查询总计和小计【英文标题】:SQL How to Query Total & Subtotal 【发布时间】:2021-04-01 08:16:00 【问题描述】:我有一个如下所示的表,其中存储了 day、order_id 和 order_type。
select day, order_id, order_type
from sample_table
day | order_id | order_type |
---|---|---|
2021-03-01 | 1 | offline |
2021-03-01 | 2 | offline |
2021-03-01 | 3 | online |
2021-03-01 | 4 | online |
2021-03-01 | 5 | offline |
2021-03-01 | 6 | offline |
2021-03-02 | 7 | online |
2021-03-02 | 8 | online |
2021-03-02 | 9 | offline |
2021-03-02 | 10 | offline |
2021-03-03 | 11 | offline |
2021-03-03 | 12 | offline |
以下是所需的输出:
day | total_order | num_offline_order | num_online_order |
---|---|---|---|
2021-03-01 | 6 | 4 | 2 |
2021-03-02 | 4 | 2 | 2 |
2021-03-03 | 2 | 2 | 0 |
有人知道如何查询以获得所需的输出吗?
【问题讨论】:
【参考方案1】:您需要对数据进行透视。在 Vertica 中实现条件聚合的一种简单方法是使用 ::
:
select day, count(*) as total_order,
sum( (order_type = 'online')::int ) as num_online,
sum( (order_type = 'offline')::int ) as num_offline
from t
group by day;
【讨论】:
我喜欢你的简洁:简单地将布尔值转换为 INT 以获得 0 与 1 的计数器,并将它们相加。【参考方案2】:使用case
和sum
:
select day,
count(1) as total_order
sum(case when order_type='offline' then 1 end) as num_offline_order,
sum(case when order_type='online' then 1 end) as num_online_order
from sample_table
group by day
order by day
【讨论】:
【参考方案3】:您还可以使用count
来聚合非空值
select
day,
count(*) as total_order,
count(case when order_type='offline' then 1 else null end) as offline_orders,
count(case when order_type='online' then 1 else null end) as online_orders
from sample_table
group by day
order by day;
【讨论】:
以上是关于SQL如何查询总计和小计的主要内容,如果未能解决你的问题,请参考以下文章