8

我有一个很长的 NVARCHAR 变量，我需要在其中替换一些这样的模式：

DECLARE @data NVARCHAR(200) = 'Hello [PAT1] stackoverflow [PAT2] world [PAT3]'

我需要[PAT%]用空格替换所有内容，如下所示：

'Hello stackoverflow world'

如何在 SQL Server 2008 中使用 T-SQL 执行此操作？

我正在搜索其他问题，我只找到了 this，但这对我没有帮助，因为我不需要保留字符串的原始部分。

4

2 回答 2

12

您可以使用此功能进行模式替换。您可以使用此SQL-Fiddle 演示对其进行测试。

CREATE FUNCTION dbo.PatternReplace
(
   @InputString VARCHAR(4000),
   @Pattern VARCHAR(100),
   @ReplaceText VARCHAR(4000)
)
RETURNS VARCHAR(4000)
AS
BEGIN
   DECLARE @Result VARCHAR(4000) SET @Result = ''
   -- First character in a match
   DECLARE @First INT
   -- Next character to start search on
   DECLARE @Next INT SET @Next = 1
   -- Length of the total string -- 8001 if @InputString is NULL
   DECLARE @Len INT SET @Len = COALESCE(LEN(@InputString), 8001)
   -- End of a pattern
   DECLARE @EndPattern INT

   WHILE (@Next <= @Len) 
   BEGIN
      SET @First = PATINDEX('%' + @Pattern + '%', SUBSTRING(@InputString, @Next, @Len))
      IF COALESCE(@First, 0) = 0 --no match - return
      BEGIN
         SET @Result = @Result + 
            CASE --return NULL, just like REPLACE, if inputs are NULL
               WHEN  @InputString IS NULL
                     OR @Pattern IS NULL
                     OR @ReplaceText IS NULL THEN NULL
               ELSE SUBSTRING(@InputString, @Next, @Len)
            END
         BREAK
      END
      ELSE
      BEGIN
         -- Concatenate characters before the match to the result
         SET @Result = @Result + SUBSTRING(@InputString, @Next, @First - 1)
         SET @Next = @Next + @First - 1

         SET @EndPattern = 1
         -- Find start of end pattern range
         WHILE PATINDEX(@Pattern, SUBSTRING(@InputString, @Next, @EndPattern)) = 0
            SET @EndPattern = @EndPattern + 1
         -- Find end of pattern range
         WHILE PATINDEX(@Pattern, SUBSTRING(@InputString, @Next, @EndPattern)) > 0
               AND @Len >= (@Next + @EndPattern - 1)
            SET @EndPattern = @EndPattern + 1

         --Either at the end of the pattern or @Next + @EndPattern = @Len
         SET @Result = @Result + @ReplaceText
         SET @Next = @Next + @EndPattern - 1
      END
   END
   RETURN(@Result)
END

资源链接。

于 2013-01-09T23:56:53.090 回答
1

+1 @Nico，我需要一个可以从字符串中删除特殊章程的函数，所以我稍微调整了你的函数以便能够做到这一点：

select dbo.PatReplaceAll('St. Martin`n tr‘an, or – in - the * field007', '[^0-9A-Z ]', '')
--Returns 'St Martinn tran or  in  the  field007'

这是功能：

CREATE FUNCTION dbo.PatReplaceAll 
(
    @Source     varchar(8000),
    @Pattern    varchar(  50),
    @Replace    varchar( 100)
)
RETURNS varchar(8000)
AS
BEGIN
    if @Source is null or @Pattern is null or @Replace is null
        return null
    if PATINDEX('%' + @Pattern + '%', @Source) = 0
        return @Source
    -- Declare the return variable here
    DECLARE @Result varchar(8000) SET @Result = ''
    -- The remainder of the @Source to work on
    DECLARE @Remainder varchar(8000) SET @Remainder = @Source
    DECLARE @Idx INT

    WHILE (LEN(@Remainder) > 0) 
    BEGIN
        SET @Idx = PATINDEX('%' + @Pattern + '%', @Remainder)
        IF @Idx = 0 --no match - return
            BEGIN
                SET @Result = @Result +  @Remainder
                BREAK
            END
        -- Concatenate characters before the match to the result
        SET @Result = @Result + SUBSTRING(@Remainder, 1, @Idx - 1)
        -- Adjust the remainder
        SET @Remainder = SUBSTRING(@Remainder, @Idx, LEN(@Remainder) + 1 - @Idx)

        SET @Idx = 1
        -- Find the last char of the pattern (aka its length)
        WHILE PATINDEX(@Pattern, SUBSTRING(@Remainder, 1, @Idx)) = 0
            SET @Idx = @Idx + 1
        --remove the pattern from the remainder
        SET @Remainder = SUBSTRING(@Remainder, @Idx + 1, LEN(@Remainder) - @Idx)
        --Add the replace string
        SET @Result = @Result + @Replace
    END
    return @Result
END
GO
于 2018-05-25T19:30:08.663 回答