Question

我有一张桌子，有一些有日期的用户...

| Employee_Name | Date       |
| CG            | 2019-04-22 |
| CG            | 2019-04-26 |
| CG            | 2019-04-24 |

我正在尝试获取结果，因此当我查询并说Date BETWEEN (@start AND @end)时，它会给我类似以下结果：

EMPLOYEE_NAME | Monday     | Tuesday | Wednesday  | Thursday | Friday     | Saturday | Sunday |
CG            | 2019-04-22 | NULL    | 2019-04-24 | NULL     | 2019-04-26 | NULL     | NULL   |

我实在不知所措地挠着头-我知道我可以用枢轴做到这一点。

Answer 1

以下内容应该可以工作（如果@ d2-@ d1 <7，则完全正确，否则它将显示每个工作日的第一个日期）：

SELECT Employee_Name
     , MIN(CASE WHEN DATENAME(dw, Date) = 'Sunday'    THEN Date END) AS Sunday
     , MIN(CASE WHEN DATENAME(dw, Date) = 'Monday'    THEN Date END) AS Monday
     , MIN(CASE WHEN DATENAME(dw, Date) = 'Tuesday'   THEN Date END) AS Tuesday
     , MIN(CASE WHEN DATENAME(dw, Date) = 'Wednesday' THEN Date END) AS Wednesday
     , MIN(CASE WHEN DATENAME(dw, Date) = 'Thursday'  THEN Date END) AS Thursday
     , MIN(CASE WHEN DATENAME(dw, Date) = 'Friday'    THEN Date END) AS Friday
     , MIN(CASE WHEN DATENAME(dw, Date) = 'Saturday'  THEN Date END) AS Saturday
FROM t
WHERE Date BETWEEN @d1 AND @d2
GROUP BY Employee_Name

Answer 2

这里是Pivot解决方案。

我添加了更多示例数据（一名员工和两名约会对象）进行测试：

declare @weekday table (weekday varchar(10))
insert into @weekday (weekday) values 
('Monday'),
('Tuesday'),
('Wednesday'),
('Thursday'),
('Friday'),
('Saturday'), 
('Sunday')

select p.employee_name,[Monday],[Tuesday],[Wednesday],[Thursday],[Friday],[Saturday],[Sunday]
from (
select t.employee_name,t.date,w.weekday from (
select *, datename(weekday,date) as weekday
from #test) t
right join @weekday w
on w.weekday = t.weekday) s
pivot
(max(date)
for weekday in ([Monday],[Tuesday],[Wednesday],[Thursday],[Friday],[Saturday],[Sunday])
) as p
where p.employee_name is not null

请相应地更改表格名称。（我使用了#test）

测试结果：

DB<>Fiddle

日期行到星期几列

2 个答案: