SQL如何按日期和多个条件聚合

Question

我有这张桌子：

ID	数量	方法	日期
01	10	一个	2022-01-24 12:27:14.440
02	80	一个	2022-01-24 12:27:14.440
01	20	D	2022-02-24 12:27:14.440
01	10	D	2022-02-24 12:27:14.440
02	20	D	2022-02-24 12:27:14.440
03	30	D	2022-02-24 12:27:14.440

我想要这个：

方法	ammount_sum_jan	n_transaction_jan	n_customer_jan	ammount_sum_feb	n_transaction_feb	n_customer_feb
一个	10	2	2	0	0	0
D	0	0	0	80	4	3

这是一个包含 7 列和行数的表，等于方法的数量。

• AMMONT：一种方法一个月内的金额总和
• N_TRANSACTIONS：一种方法一个月内的交易数量
• N_CUSTOMER：一个月内使用该方法的客户（id）数

我可以通过一个查询得到它吗？

Answer 1

您希望按方法汇总数据，并为 1 月数据和 2 月数据设置单独的列。 您可以通过条件聚合（聚合函数中的CASE表达式）得到它，

select
  method,
  sum(case when month(date) = 1 then ammount else 0 end) as ammount_sum_jan,
  count(case when month(date) = 1 then 1 end) as n_transaction_jan,
  count(distinct case when month(date) = 1 then id end) as n_customer_jan,
  sum(case when month(date) = 2 then ammount else 0 end) as ammount_sum_feb,
  count(case when month(date) = 2 then 1 end) as n_transaction_feb,
  count(distinct case when month(date) = 2 then id end) as n_customer_feb
from mytable
group by method
order by method;

Answer 2

它被称为枢轴，并且对于 dfix 日期看起来像这样。 年份和月份方法的聚合，您可以对您的号码进行计数或求和

select 
  `method`,
  SUM(CASE WHEN EXTRACT( YEAR_MONTH FROM `date` ) = '202201' then `ammount` ELSe 0 END) ammount_sum_jan,
  SUM(CASE WHEN EXTRACT( YEAR_MONTH FROM `date` ) = '202201' then 1 ELSe 0 END) n_transaction_jan,
  COUNT(DISTINCT CASE WHEN EXTRACT( YEAR_MONTH FROM `date` ) = '202201' then `d` ELSe 0 END)    n_customer_jan,
  SUM(CASE WHEN EXTRACT( YEAR_MONTH FROM `date` ) = '202202' then `ammount` ELSe 0 END) ammount_sum_feb,
  SUM(CASE WHEN EXTRACT( YEAR_MONTH FROM `date` ) = '202202' then 1 ELSe 0 END) n_transaction_feb,
  COUNT(DISTINCT CASE WHEN EXTRACT( YEAR_MONTH FROM `date` ) = '202202' then `d` ELSe 0 END)    n_customer_feb
from tab1
GROUP BY `method`

http://www.sqlfiddle.com/#!9/31d8ef/10

更有趣的是让这种动态