【问题标题】:SQL remove multiple time stamp from any part of a stringSQL从字符串的任何部分删除多个时间戳
【发布时间】:2018-01-31 01:56:18
【问题描述】:

我有一个包含多个日期时间戳的文本字段,我只需要删除时间戳。此字段中还有带有回车的文本。我想维护文本、日期和回车。

任何人都有这方面的经验并且可以分享选择来完成这个吗?

这是文本字段的示例:

[10/19/2015 5:24:02 PM Tech1] Repaired heaters [10/21/2015 8:36:28 AM Tech1] CHECKED ALL HEATER OPERATION  

任何见解将不胜感激。提前感谢您的帮助!

【问题讨论】:

  • 总是 2 个时间戳?
  • 并非总是如此。可能超过 2 个
  • 日期/时间格式是绝对一致还是有多种格式,例如24 小时制,星期几,...?
  • 提示...拆分字符串函数然后子字符串....然后东西
  • 日期时间格式一致

标签: sql-server tsql datetime


【解决方案1】:

这是一种使用STUFF 和拆分器功能的方法,同时考虑了一位数和两位数的小时数。

declare @table table (vars varchar(4000))
insert into @table
values
('[10/19/2015 5:24:02 PM Tech1] Repaired heaters' + char(10) + '[10/21/2015 12:36:28 AM Tech1] CHECKED ALL HEATER OPERATION' + char(13) + '[10/22/2015 3:36:28 PM Tech1] somethingelse')
,('[11/19/2015 5:24:02 PM Tech1] asdfsdaf  [11/21/2015 8:36:28 AM Tech1] dddddddddddd [11/22/2015 3:36:28 PM Tech1] yyyyyyyyy')
,('[1/19/2015 11:24:02 PM Tech1] here went something  [09/21/2015 08:36:28 AM Tech1] Tech 1 did something else[11/22/2015 10:36:28 PM Tech1] last')

;with cte as(
select
    vars
    ,RN = dense_rank() over (order by vars)
    --,Item
    ,Item = '[' + stuff(Item,charindex(' ',Item), 3 + case when charindex(' AM ',Item) = 0 then charindex(' PM ',Item) else charindex(' AM ',Item) end  - charindex(' ',Item),'')
from
    @table
    cross apply
    dbo.DelimitedSplit8K(vars,'[')
where
    Item <>'')



select distinct
    vars
    ,RemovedTime =
                STUFF((
                      SELECT ' ' + c2.Item
                      FROM cte c2
                      WHERE c.RN = c2.RN
                      FOR XML PATH(''), TYPE).value('.', 'NVARCHAR(MAX)'), 1, 1, '')
from
    cte c

拆分器可以在Jeff Moden's Article找到

CREATE FUNCTION [dbo].[DelimitedSplit8K] (@pString VARCHAR(8000), @pDelimiter CHAR(1))
--WARNING!!! DO NOT USE MAX DATA-TYPES HERE!  IT WILL KILL PERFORMANCE!

RETURNS TABLE WITH SCHEMABINDING AS
RETURN

/* "Inline" CTE Driven "Tally Table" produces values from 1 up to 10,000...
enough to cover VARCHAR(8000)*/

  WITH E1(N) AS (
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
                ),                          --10E+1 or 10 rows
       E2(N) AS (SELECT 1 FROM E1 a, E1 b), --10E+2 or 100 rows
       E4(N) AS (SELECT 1 FROM E2 a, E2 b), --10E+4 or 10,000 rows max
 cteTally(N) AS (--==== This provides the "base" CTE and limits the number of rows right up front
                     -- for both a performance gain and prevention of accidental "overruns"
                 SELECT TOP (ISNULL(DATALENGTH(@pString),0)) ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) FROM E4
                ),
cteStart(N1) AS (--==== This returns N+1 (starting position of each "element" just once for each delimiter)
                 SELECT 1 UNION ALL
                 SELECT t.N+1 FROM cteTally t WHERE SUBSTRING(@pString,t.N,1) = @pDelimiter
                ),
cteLen(N1,L1) AS(--==== Return start and length (for use in substring)
                 SELECT s.N1,
                        ISNULL(NULLIF(CHARINDEX(@pDelimiter,@pString,s.N1),0)-s.N1,8000)
                   FROM cteStart s
                )
--===== Do the actual split. The ISNULL/NULLIF combo handles the length for the final element when no delimiter is found.
 SELECT ItemNumber = ROW_NUMBER() OVER(ORDER BY l.N1),
        Item       = SUBSTRING(@pString, l.N1, l.L1)
   FROM cteLen l
;
GO

【讨论】:

    【解决方案2】:

    更新:误解了要求,根据下面的 cmets 更新了我的解决方案。

    为此,您需要PatternSplitCM

    declare @table table (someid int identity, somestring varchar(1000));
    insert @table(somestring) values 
    ('[10/19/2015 5:24:02 PM Tech1] Repaired heaters [10/21/2015 8:36:28 AM Tech1] CHECKED ALL HEATER OPERATION'),
    ('[02/28/2015 5:24:02 PM Tech1] Repaired more stuff [12/01/2015 2:36:28 AM Tech1] CHECKED ALL HEATER OPERATION');
    
    with parseMe as
    (
      select 
        someid, 
        ItemNumber,
        Item = case 
               when Item like '[0-9]:[0-9][0-9]:[0-9][0-9]' then '<exclude>' 
               else replace(replace(item, ' AM ',''), ' PM ', '') end
      from @table t
      cross apply dbo.PatternSplitCM(t.somestring, '[0-9:/]')
    )
    select
      someid,
      newString = 
        replace((
          select item+''
          from parseMe p2 
            where p1.someid = p2.someid 
          order by ItemNumber
          for xml path(''), type
        ).value('.', 'varchar(1000)'), '<exclude>','')
    from parseMe p1
    group by someid;
    

    【讨论】:

    • 这会替换整个日期时间戳,而不仅仅是时间部分。
    • 如果字符串中的其他任何地方有大括号,那不是时间戳的一部分,它也会删除它们......
    • @scsimon 我根据你的 cmets 更新了我的解决方案
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