Question

我的字符串值为07052018080504623

它代表MM DD YYYY HH MM SS MS

表示

MM = 07, DD = 05, YYYY=2018, HH = 08, MM=05 , SS= 04, MS = 623

现在我有一个定义为

的表

declare @t table (WorkRequestId varchar(100))
insert into @t values
('07052018080504623'),('07062018012756663'),('07062018020148130'),('07062018095201231'),
('07062018102203805'),('07062018103718059'),('07062018110304836'),('07062018115356135'),
('07062018120624983'),('07062018124035480'),('07062018080504623'),('07062018070504623')

select 
    *   
from @t

记录应按升序排序，以使

WorkRequestId

07052018080504623
07062018095201231
07062018102203805
07062018103718059
07062018110304836
07062018115356135
07062018120624983
07062018124035480
07062018012756663
07062018020148130

截止时间为前一天的晚上8点至次日的19:59:59。在我们的示例中，05是前一天，而06是第二天。

第二天凌晨1点至7:59:59之间也没有交易发生。从第二天的上午8点再次开始，一直持续到 19:59:59 PM。

因此，当我们遇到07062018012756663, the DD=06 and HH = 01时。这意味着13HRs (i.e. 1PM) of 6th。与07062018020148130 where DD=06 and HH = 02 (i.e. 2PM or 14Hrs).

相同

但是07062018095201231 where DD=06 and HH = 09 means 9AM of 6th.

这就是为什么

07062018095201231 comes before 07062018012756663 and 07062018020148130

订购时

到目前为止，我的尝试（尚不正确）

select 
    *       
    ,DY=SUBSTRING([WorkRequestId],3,2)
    ,HH = SUBSTRING([WorkRequestId],9,2)
    ,CurrentDY=CONVERT(varchar(2), getdate(), 103)
from @t

order by left([WorkRequestId],8) + 
    cast(iif( 
            SUBSTRING([WorkRequestId],3,2) = '6',--CONVERT(varchar(2), getdate(), 103), 
            iif(cast(SUBSTRING([WorkRequestId],9,2) as int) between 1 and 7,
            cast(SUBSTRING([WorkRequestId],9,2) as int)+12,SUBSTRING([WorkRequestId],9,2)),
            cast(SUBSTRING([WorkRequestId],9,2)as varchar(4)))as varchar(20))
    +right([WorkRequestId],7)

Answer 1

因此，我想您现在应该意识到，将日期存储为字符串（实际上，将任何内容存储在错误的数据类型中）是一种不好的做法。
正确的解决方案是更改数据库结构，以将数据保留为DateTime2而不是字符串。但是，假设由于某种原因无法执行此操作，则可以通过将字符串值转换为datetime2，在小时介于1 am和8 am之间的情况下，添加12小时，然后按此排序，来获得所需的结果日期。
我以繁琐的方式写了我的建议，因为我想展示过程的每个部分-虽然可以在单个查询中完成，但我使用了3个通用表表达式-再次，这只是为了说明解决方案的每个步骤：

;WITH CTEDateParts AS -- break down the string to it's parts
(
    SELECT  WorkRequestId, 
            SUBSTRING(WorkRequestId, 5, 4) As Year,
            SUBSTRING(WorkRequestId, 1, 2) As Month,
            SUBSTRING(WorkRequestId, 3, 2) As Day,
            SUBSTRING(WorkRequestId, 9, 2) As Hour,
            SUBSTRING(WorkRequestId, 11, 2) As Minute,
            SUBSTRING(WorkRequestId, 13, 2) As Second,
            SUBSTRING(WorkRequestId, 15, 3) As Millisecond
    FROM @t
), CTEDates AS -- create datetime values from the string parts
(
    SELECT  WorkRequestId, 
            CAST(Year +'-'+ Month +'-'+ Day +'T'+ 
                 Hour +':'+ Minute +':'+ Second +'.'+ Millisecond As DateTime2(7)) As DateValue
    FROM CTEDateParts
), CTEFixedDates AS -- add 12 hours for hours between 1 and 8 a.m.
(
    SELECT  WorkRequestId, 
            DateValue,
            CASE WHEN DATEPART(HOUR, DateValue) >= 1 AND DATEPART(HOUR, DateValue) <= 8 THEN 
                DATEADD(Hour, 12, DateValue)
            ELSE
                DateValue
            END As FixedDate 
    FROM CTEDates
)

-- finally, select order by the FixedDate column
SELECT WorkRequestId
FROM CTEFixedDates
ORDER BY FixedDate

结果：

WorkRequestId
07052018080504623
07062018095201231
07062018102203805
07062018103718059
07062018110304836
07062018115356135
07062018120624983
07062018124035480
07062018012756663
07062018020148130
07062018070504623
07062018080504623

Answer 2

您可以尝试以下操作：

select 
    *
from @t 
order by left (WorkRequestId, 8) + (case when SUBSTRING(WorkRequestId, 9,2) between '01' and '07' then CAST(SUBSTRING(WorkRequestId, 9,2) + 12 AS CHAR(2)) else SUBSTRING(WorkRequestId, 9,2) end) +  SUBSTRING(WorkRequestId, 11,7)

如何按日和小时字段排序？

2 个答案: