仅反转sql server中字符串的数字部分

Question

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

Answer 1

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

Answer 2

只需使用PATINDEX进行搜索，按部分追加到结果字符串：

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

测试SQL

declare @Values table (Value varchar(20))
INSERT @Values VALUES
('Test123Hello'),
('Tt143 Hello'),
('12Hll'), 
('Tt123H3451end'),
(''),
(NULL)

SELECT Value, dbo.fn_ReverseDigits(Value) ReversedValue FROM @Values

结果

Value                ReversedValue
-------------------- --------------------
Test123Hello         Test321Hello
Tt143 Hello          Tt341 Hello
12Hll                21Hll
Tt123H3451end        Tt321H1543end

NULL                 NULL

Answer 3

希望这个帮助：

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)

Answer 4

使用此功能，它也可以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，那么首先创建它我已经包含在下面

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

Answer 5

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('')), '&#x20;',' ') AS col2
FROM mytable AS m

使用计数表以“取消”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。

小提琴演示在这里

Answer 6

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)

从剩下的值来看，你需要为你得到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，如果将值（之前和之后）放入表[B＆A]并在'之前'和'之后'调用列，则运行以下命令：

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'Test321Hello'。

但是，除非您为每个字母数字形状创建类似的“[0-9] [az]”类型模板并将其应用于[B＆A]表，否则其他3行不会受到影响。 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[az]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。