gpt4 book ai didi

sql - 从位掩码确定一周中的几天

转载 作者:行者123 更新时间:2023-12-03 01:51:14 25 4
gpt4 key购买 nike

我正在使用第三方应用程序,并尝试根据数据提取有关类次信息的有意义的信息。

shift_pattern_start_dt     pattern
2014-05-27 1111000
2015-10-25 1110011

2014-05-27 是一个Tuesday,该形态的开始位置是Tuesday。因此,我希望结果显示星期二星期三星期四星期五

2015-10-25 是一个星期日,该形态的起始位置是星期日。结果应为星期日星期一星期二星期五星期六

对于确定正确的工作日有什么想法或建议吗?

最佳答案

Declare @YourTable table (shift_pattern_start_dt date, pattern varchar(25))
Insert Into @YourTable values
('2014-05-27','1111000'),
('2015-10-25','1110011')

Select *
,NewCol = concat(
IIF(substring(pattern,1,1)='1', +DateName(WEEKDAY,shift_pattern_start_dt),'')
,IIF(substring(pattern,2,1)='1',','+DateName(WEEKDAY,dateadd(DAY,1,shift_pattern_start_dt)),null)
,IIF(substring(pattern,3,1)='1',','+DateName(WEEKDAY,dateadd(DAY,2,shift_pattern_start_dt)),null)
,IIF(substring(pattern,4,1)='1',','+DateName(WEEKDAY,dateadd(DAY,3,shift_pattern_start_dt)),null)
,IIF(substring(pattern,5,1)='1',','+DateName(WEEKDAY,dateadd(DAY,4,shift_pattern_start_dt)),null)
,IIF(substring(pattern,6,1)='1',','+DateName(WEEKDAY,dateadd(DAY,5,shift_pattern_start_dt)),null)
,IIF(substring(pattern,7,1)='1',','+DateName(WEEKDAY,dateadd(DAY,6,shift_pattern_start_dt)),null)
)
From @YourTable

返回

shift_pattern_start_dt  pattern   NewCol
2014-05-27 1111000 Tuesday,Wednesday,Thursday,Friday
2015-10-25 1110011 Sunday,Monday,Tuesday,Friday,Saturday

EDIT - Cross Apply Version

Select A.*
,B.*
From @YourTable A
Cross Apply (
Select NewCol =Stuff((Select ',' +D
From (
Select N,D = IIF(substring(A.pattern,N,1)='0',null,DateName(WEEKDAY,DateAdd(DAY,N-1,A.shift_pattern_start_dt)))
From (Values (1),(2),(3),(4),(5),(6),(7)) N(N)
) B1
For XML Path ('')),1,1,'')
) B

Execution Plan for Concat() Approach

enter image description here

Execution Plan for Cross Apply Approach

enter image description here

关于sql - 从位掩码确定一周中的几天,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42864473/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com