使用T-SQL，我试图找到最简单的方法来反转字符串中的数字。所以像Test123Hello这样的字符串有Test321Hello。
[Before] [After]
Test123Hello Test321Hello
Tt143 Hello Tt341 Hello
12Hll 21Hll
Tt123H3451end Tt321H1543end
答案 0 :(得分：3)
你可以使用这个功能
CREATE FUNCTION [dbo].[fn_ReverseDigit_MA]
(
@Str_IN nVARCHAR(max)
)
RETURNS NVARCHAR(max)
AS
BEGIN
DECLARE @lenstr AS INT =LEN(@Str_IN)
DECLARE @lastdigend AS INT=0
while (@lastdigend<@lenstr)
BEGIN
DECLARE @strPart1 AS NVARCHAR(MAX)=LEFT(@Str_IN,@lastdigend)
declare @lenstrPart1 AS INT=LEN(@strPart1)
DECLARE @strPart2 AS NVARCHAR(MAX)=RIGHT(@Str_IN,@lenstr-@lastdigend)
declare @digidx as int=patindex(N'%[0-9]%' ,@strPart2)+@lenstrPart1
IF(@digidx=@lenstrPart1)
BEGIN
BREAK;
END
DECLARE @strStartdig AS NVARCHAR(MAX) = RIGHT(@Str_IN,@lenstr-@digidx+1)
declare @NDidx as int=patindex(N'%[^0-9]%' ,@strStartdig)+@digidx-1
IF(@NDidx<=@digidx)
BEGIN
SET @NDidx=@lenstr+1
END
DECLARE @strRet AS NVARCHAR(MAX)=LEFT(@Str_IN,@digidx-1) +REVERSE(SUBSTRING(@Str_IN,@digidx,@NDidx-@digidx)) +RIGHT(@Str_IN,@lenstr-@NDidx+1)
SET @Str_IN=@strRet
SET @lastdigend=@NDidx-1
END
return @Str_IN
END
答案 1 :(得分：2)
只需使用PATINDEX进行搜索，按部分追加结果字符串：
Sub StartsWithA()
Cells(1, Columns.Count).End(xlToLeft).Offset(1, 1).Resize(Range("A" & Rows.Count).End(xlUp).Row - 1, 1) = "=IF(UPPER(LEFT(A2,1))=""A"",""pass"",""fail"")"
Cells(1, Columns.Count).End(xlToLeft).Offset(0, 1).Formula = "Result"
activesheet.calculate 'In case your calcs are set to manual
Cells(1, Columns.Count).End(xlToLeft).Offset(0, 0).Resize(Range("A" & Rows.Count).End(xlUp).Row, 1).Copy
Cells(1, Columns.Count).End(xlToLeft).Offset(0, 0).Resize(Range("A" & Rows.Count).End(xlUp).Row, 1).PasteSpecial xlPasteValues
Application.CutCopyMode = False
End Sub
测试SQL
CREATE FUNCTION [dbo].[fn_ReverseDigits]
(
@Value nvarchar(max)
)
RETURNS NVARCHAR(max)
AS
BEGIN
IF @Value IS NULL
RETURN NULL
DECLARE
@TextIndex int = PATINDEX('%[^0-9]%', @Value),
@NumIndex int = PATINDEX('%[0-9]%', @Value),
@ResultValue nvarchar(max) = ''
WHILE LEN(@ResultValue) < LEN(@Value)
BEGIN
-- Set the index to end of the string if the index is 0
SELECT @TextIndex = CASE WHEN @TextIndex = 0 THEN LEN(@Value) + 1 ELSE LEN(@ResultValue) + @TextIndex END
SELECT @NumIndex = CASE WHEN @NumIndex = 0 THEN LEN(@Value) + 1 ELSE LEN(@ResultValue) + @NumIndex END
IF @NumIndex < @TextIndex
SELECT @ResultValue = @ResultValue + REVERSE(SUBSTRING(@Value, @NumIndex, @TextIndex -@NumIndex))
ELSE
SELECT @ResultValue = @ResultValue + (SUBSTRING(@Value, @TextIndex, @NumIndex - @TextIndex))
-- Update index variables
SELECT
@TextIndex = PATINDEX('%[^0-9]%', SUBSTRING(@Value, LEN(@ResultValue) + 1, LEN(@Value) - LEN(@ResultValue))),
@NumIndex = PATINDEX('%[0-9]%', SUBSTRING(@Value, LEN(@ResultValue) + 1, LEN(@Value) - LEN(@ResultValue)))
END
RETURN @ResultValue
END
结果
declare @Values table (Value varchar(20))
INSERT @Values VALUES
('Test123Hello'),
('Tt143 Hello'),
('12Hll'),
('Tt123H3451end'),
(''),
(NULL)
SELECT Value, dbo.fn_ReverseDigits(Value) ReversedValue FROM @Values
答案 2 :(得分：1)
declare @s nvarchar(128) ='Test321Hello'
declare @numStart as int, @numEnd as int
select @numStart =patindex('%[0-9]%',@s)
select @numEnd=len(@s)-patindex('%[0-9]%',REVERSE(@s))
select
SUBSTRING(@s,0,@numstart)+
reverse(SUBSTRING(@s,@numstart,@numend-@numstart+2))+
SUBSTRING(@s,@numend+2,len(@s)-@numend)
答案 3 :(得分：1)
使用此功能，它也会处理多次出现的数字
create FUNCTION [dbo].[GetReverseNumberFromString] (@String VARCHAR(2000))
RETURNS VARCHAR(1000)
AS
BEGIN
DECLARE @Count INT
DECLARE @IntNumbers VARCHAR(1000)
declare @returnstring varchar(max)=@String;
SET @Count = 0
SET @IntNumbers = ''
WHILE @Count <= LEN(@String)
BEGIN
IF SUBSTRING(@String, @Count, 1) >= '0'
AND SUBSTRING(@String, @Count, 1) <= '9'
BEGIN
SET @IntNumbers = @IntNumbers + SUBSTRING(@String, @Count, 1)
END
IF (
SUBSTRING(@String, @Count + 1, 1) < '0'
OR SUBSTRING(@String, @Count + 1, 1) > '9'
)
AND SUBSTRING(@String, @Count, 1) >= '0'
AND SUBSTRING(@String, @Count, 1) <= '9'
BEGIN
SET @IntNumbers = @IntNumbers + ','
END
SET @Count = @Count + 1
END
declare @RevStrings table (itemz varchar(50))
INSERT INTO @RevStrings(itemz)
select items from dbo.Split(@IntNumbers,',')
select @returnstring = Replace(@returnstring, itemz,REVERSE(itemz))from @RevStrings
RETURN @returnstring
END
你的样本字符串
select [dbo].[GetReverseNumberFromString]('Tt123H3451end')
结果
Tt321H1543end
更新：
如果你没有Split功能，那么首先创建它 我把它包括在下面
create FUNCTION Split
(
@Input NVARCHAR(MAX),
@Character CHAR(1)
)
RETURNS @Output TABLE (
Items NVARCHAR(1000)
)
AS
BEGIN
DECLARE @StartIndex INT, @EndIndex INT
SET @StartIndex = 1
IF SUBSTRING(@Input, LEN(@Input) - 1, LEN(@Input)) <> @Character
BEGIN
SET @Input = @Input + @Character
END
WHILE CHARINDEX(@Character, @Input) > 0
BEGIN
SET @EndIndex = CHARINDEX(@Character, @Input)
INSERT INTO @Output(Items)
SELECT SUBSTRING(@Input, @StartIndex, @EndIndex - 1)
SET @Input = SUBSTRING(@Input, @EndIndex + 1, LEN(@Input))
END
RETURN
END
GO
答案 4 :(得分：1)
这是一种基于集合的方法：
;WITH Tally (n) AS
(
SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL))
FROM (VALUES(0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) a(n)
CROSS JOIN (VALUES(0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) b(n)
CROSS JOIN (VALUES(0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) c(n)
), UnpivotCTE AS (
SELECT id, x.c, n, y.isNumber,
n - ROW_NUMBER() OVER (PARTITION BY id, y.isNumber
ORDER BY n) AS grp
FROM mytable
CROSS JOIN Tally
CROSS APPLY (SELECT SUBSTRING(col, n, 1)) AS x(c)
CROSS APPLY (SELECT ISNUMERIC(x.c)) AS y(isNumber)
WHERE n <= LEN(col)
), ToConcatCTE AS (
SELECT id, c, n, isNumber,
grp + MIN(n) OVER (PARTITION BY id, isNumber, grp) AS grpAsc
FROM UnpivotCTE
)
SELECT id, col,
REPLACE(
(SELECT c AS [text()]
FROM ToConcatCTE AS t
WHERE t.id = m.id
ORDER BY id,
grpAsc,
CASE WHEN isNumber = 0 THEN n END,
CASE WHEN isNumber = 1 THEN n END DESC
FOR XML PATH('')), ' ',' ') AS col2
FROM mytable AS m
使用计数表以便“忽略”＆＃39;字符串的所有字符。然后使用ROW_NUMBER来识别数字和非数字字符的孤岛。最后，FOR XML PATH用于重建具有反转的数字岛的初始字符串：ORDER BY用于按相反的顺序对数字字符的岛进行排序。
答案 5 :(得分：0)
这将执行您要求的特定字符串：
select
substring('Test123Hello',1,4)
+
reverse(substring('Test123Hello',5,3))
+
substring('Test123Hello',8,5)
根据其余的值来判断，您需要为您获得的任何字母数字模式制作模板。例如，您可以将上述内容应用于具有以下形状的任何值：
select * from [B&A] where [before] like '[a-z][a-z][a-z][a-z][0-9][0-9][0-9]
[a-z][a-z][a-z][a-z][a-z]'
换句话说，如果您将值（之前和之后）放入表[B＆amp; A]并在之前调用列＆＃39;和＆＃39;之后＆＃39;然后跑了这个：
select
substring(before,1,4)
+
reverse(substring(before,5,3))
+
substring(before,8,5) as [after]
from [B&A] where [before] like '[a-z][a-z][a-z][a-z][0-9][0-9][0-9][a-z]
[a-z][a-z][a-z][a-z]'
然后它会给你＆＃39; Test321Hello＆＃39;。
但是除非您创建了类似的行，否则其他3行不会受到影响 ＆＃39; [0-9] [A-Z]＆＃39;为每个字母数字形状键入模板，并将其应用于[B＆amp; A]表。您必须将结果选择到临时表或另一个表中。
通过依次应用每个模板，您可以获得大部分内容，然后您必须查看有多少行未受影响，并检查字母数字形状是什么，并制作更多模板。最终你有一组代码，如果你运行它会捕获所有可能的组合。
您可以坐下来以这种方式设计代码，捕获[a-z]和[0-9]的所有可能组合。很大程度上取决于您正在处理的最大字符数。