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
天 | order_id | 订单类型 |
---|---|---|
2021-03-01 | 1 | 离线 |
2021-03-01 | 2 | 离线 |
2021-03-01 | 3 | 在线 |
2021-03-01 | 4 | 在线 |
2021-03-01 | 5 | 离线 |
2021-03-01 | 6 | 离线 |
2021-03-02 | 7 | 在线 |
2021-03-02 | 8 | 在线 |
2021-03-02 | 9 | 离线 |
2021-03-02 | 10 | 离线 |
2021-03-03 | 11 | 离线 |
2021-03-03 | 12 | 离线 |
以下是所需的输出:
天 | 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?
答案 0 :(得分:2)
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;
答案 1 :(得分:0)
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
答案 2 :(得分:0)
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;