Question

来自印度尼西亚的Fajar，我有问题用sql server创建报告。我有一个交易表，如下所示，

Date       | Trx_ID| Member_id| Invt_id |Branch_id
01-01-2013 | tr001 | 0002     | invt002 |br001
01-01-2013 | tr001 | 0002     | invt003 |br001
01-01-2013 | tr001 | 0002     | invt004 |br001
01-03-2013 | tr002 | 0001     | invt001 |br001
01-03-2013 | tr002 | 0001     | invt003 |br001
01-04-2013 | tr003 | 0003     | invt002 |br001
01-04-2013 | tr003 | 0003     | invt004 |br001
01-05-2013 | tr004 | 0002     | invt005 |br001
01-05-2013 | tr004 | 0002     | invt006 |br001
01-05-2013 | tr005 | 0005     | invt004 |br001
01-05-2013 | tr005 | 0005     | invt005 |br001
01-05-2013 | tr005 | 0005     | invt006 |br001
01-06-2013 | tr006 | 0001     | invt001 |br001
01-06-2013 | tr006 | 0001     | invt002 |br001

我希望像这样创建报告

Branch_id| weekdays | weekend|weekdays_Weekend
br001    |  2       | 1      | 2

由member_id表示的平均组。如果在周末进行交易的会员在工作日计算，如果在工作日进行交易的会员在周末也计算它，但如果在平日和周末进行交易的会员在工作日_周末计算它。周末是：周六和周日工作日是：周一，周二，周三，周四和周五 weekdays_weekend如果成员执行的交易超过1并且在周末和工作日进行

当我尝试

时这是错误的查询

select
  max(branch_id)
, count(DISTINCT trx_id)
, sum(case when datename(dw,date) in ('Monday','Tuesday','Wednesday','Thursday','Friday') then f.trx_id else null end) AS 'WeekDays'
, sum(case when datename(dw,date) in ('Saturday', 'Sunday') then 1 else 0 end) as 'Weekend'
, sum(case when datename(dw,date) in ('Monday','Tuesday','Wednesday','Thursday','Friday','Saturday', 'Sunday') then 1 else 0 end) as 'Weekdays_weekend'
from Transaction

where branch_id = '14067' and convert (char(6), date, 112) = '201301'
group by branch_id

结果不是包含trx_id和member_id

的组

Branch_id|weekdays|weekend|weekdays_weekend
br001    | 7      | 7     | 0

如果我的英语不好，我很抱歉请帮我创建报告

Answer 1

嗯，我不能通过计算工作日_周末来了解你的情况，但是bewlow是查询周末或工作日的总交易

SELECT branch_id,
       SUM(CASE WHEN DATEPART(dw,[Date]) NOT IN (1,7) THEN 1 ELSE 0 END) as weekdays,
       SUM(CASE WHEN DATEPART(dw,[Date])  IN (1,7) THEN 1 ELSE 0 END) as weekend
FROM tableName
GROUP BY branch_id

DATEPART（）函数提取dw传递给它的星期几。返回的值是1到7，星期六是7，星期日是1

sql获取日期的名称并计算它

1 个答案: