14

我需要对存储 URL 编码文本的旧表运行查询。我需要在我的结果中解码此文本。我如何实现这一目标?

4

4 回答 4

29

尝试其中之一:

CREATE FUNCTION dbo.UrlDecode(@url varchar(3072))
RETURNS varchar(3072)
AS
BEGIN 
    DECLARE @count int, @c char(1), @cenc char(2), @i int, @urlReturn varchar(3072) 
    SET @count = Len(@url) 
    SET @i = 1 
    SET @urlReturn = '' 
    WHILE (@i <= @count) 
     BEGIN 
        SET @c = substring(@url, @i, 1) 
        IF @c LIKE '[!%]' ESCAPE '!' 
         BEGIN 
            SET @cenc = substring(@url, @i + 1, 2) 
            SET @c = CHAR(CASE WHEN SUBSTRING(@cenc, 1, 1) LIKE '[0-9]' 
                                THEN CAST(SUBSTRING(@cenc, 1, 1) as int) 
                                ELSE CAST(ASCII(UPPER(SUBSTRING(@cenc, 1, 1)))-55 as int) 
                            END * 16 + 
                            CASE WHEN SUBSTRING(@cenc, 2, 1) LIKE '[0-9]' 
                                THEN CAST(SUBSTRING(@cenc, 2, 1) as int) 
                                ELSE CAST(ASCII(UPPER(SUBSTRING(@cenc, 2, 1)))-55 as int) 
                            END) 
            SET @urlReturn = @urlReturn + @c 
            SET @i = @i + 2 
         END 
        ELSE 
         BEGIN 
            SET @urlReturn = @urlReturn + @c 
         END 
        SET @i = @i +1 
     END 
    RETURN @urlReturn
END
GO

来自http://sqlblog.com/blogs/peter_debetta/archive/2007/03/09/t-sql-urldecode.aspx


CREATE FUNCTION dbo.fnDeURL
(
    @URL VARCHAR(8000)
)
RETURNS VARCHAR(8000)
AS
BEGIN
    DECLARE @Position INT,
        @Base CHAR(16),
        @High TINYINT,
        @Low TINYINT,
        @Pattern CHAR(21)

    SELECT  @Base = '0123456789abcdef',
        @Pattern = '%[%][0-9a-f][0-9a-f]%',
        @URL = REPLACE(@URL, '+', ' '),
        @Position = PATINDEX(@Pattern, @URL)

    WHILE @Position > 0
        SELECT  @High = CHARINDEX(SUBSTRING(@URL, @Position + 1, 1), @Base COLLATE Latin1_General_CI_AS),
            @Low = CHARINDEX(SUBSTRING(@URL, @Position + 2, 1), @Base COLLATE Latin1_General_CI_AS),
            @URL = STUFF(@URL, @Position, 3, CHAR(16 * @High + @Low - 17)),
            @Position = PATINDEX(@Pattern, @URL)

    RETURN  @URL
END

来自http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=88926

于 2010-09-30T17:36:08.993 回答
3

如果您需要支持非英文字符(unicode)的解决方案,CodeProject上有一个很好的示例:

CREATE FUNCTION [dbo].[UrlDecode] (
    @URL NVARCHAR(4000) )   RETURNS NVARCHAR(4000) AS BEGIN
    DECLARE @Position INT, @Base CHAR(16), @High TINYINT, @Low TINYINT, @Pattern CHAR(21)
    DECLARE @Byte1Value INT, @SurrogateHign INT, @SurrogateLow INT
    SELECT @Pattern = '%[%][0-9a-f][0-9a-f]%', @Position = PATINDEX(@Pattern, @URL)

    WHILE @Position > 0
    BEGIN
       SELECT @High = ASCII(UPPER(SUBSTRING(@URL, @Position + 1, 1))) - 48,
              @Low  = ASCII(UPPER(SUBSTRING(@URL, @Position + 2, 1))) - 48,
              @High = @High / 17 * 10 + @High % 17,
              @Low  = @Low  / 17 * 10 + @Low  % 17,
              @Byte1Value = 16 * @High + @Low
       IF @Byte1Value < 128 --1-byte UTF-8
          SELECT @URL = STUFF(@URL, @Position, 3, NCHAR(@Byte1Value)),
                 @Position = PATINDEX(@Pattern, @URL)
       ELSE IF @Byte1Value >= 192 AND @Byte1Value < 224 AND @Position > 0 --2-byte UTF-8
       BEGIN
           SELECT @Byte1Value = (@Byte1Value & (POWER(2,5) - 1)) * POWER(2,6),
                  @URL = STUFF(@URL, @Position, 3, ''),
                  @Position = PATINDEX(@Pattern, @URL)
           IF @Position > 0
              SELECT @High = ASCII(UPPER(SUBSTRING(@URL, @Position + 1, 1))) - 48,
                     @Low  = ASCII(UPPER(SUBSTRING(@URL, @Position + 2, 1))) - 48,
                     @High = @High / 17 * 10 + @High % 17,
                     @Low  = @Low  / 17 * 10 + @Low  % 17,
                     @Byte1Value = @Byte1Value + ((16 * @High + @Low) & (POWER(2,6) - 1)),
                     @URL = STUFF(@URL, @Position, 3, NCHAR(@Byte1Value)),
                     @Position = PATINDEX(@Pattern, @URL)
       END
       ELSE IF @Byte1Value >= 224 AND @Byte1Value < 240 AND @Position > 0 --3-byte UTF-8
       BEGIN
           SELECT @Byte1Value = (@Byte1Value & (POWER(2,4) - 1)) * POWER(2,12),
                  @URL = STUFF(@URL, @Position, 3, ''),
                  @Position = PATINDEX(@Pattern, @URL)
           IF @Position > 0
              SELECT @High = ASCII(UPPER(SUBSTRING(@URL, @Position + 1, 1))) - 48,
                     @Low  = ASCII(UPPER(SUBSTRING(@URL, @Position + 2, 1))) - 48,
                     @High = @High / 17 * 10 + @High % 17,
                     @Low  = @Low  / 17 * 10 + @Low  % 17,
                     @Byte1Value = @Byte1Value + ((16 * @High + @Low) & (POWER(2,6) - 1)) * POWER(2,6),
                     @URL = STUFF(@URL, @Position, 3, ''),
                     @Position = PATINDEX(@Pattern, @URL)
           IF @Position > 0
              SELECT @High = ASCII(UPPER(SUBSTRING(@URL, @Position + 1, 1))) - 48,
                     @Low  = ASCII(UPPER(SUBSTRING(@URL, @Position + 2, 1))) - 48,
                     @High = @High / 17 * 10 + @High % 17,
                     @Low  = @Low  / 17 * 10 + @Low  % 17,
                     @Byte1Value = @Byte1Value + ((16 * @High + @Low) & (POWER(2,6) - 1)),
                     @URL = STUFF(@URL, @Position, 3, NCHAR(@Byte1Value)),
                     @Position = PATINDEX(@Pattern, @URL)
       END
       ELSE IF @Byte1Value >= 240 AND @Position > 0  --4-byte UTF-8
       BEGIN
           SELECT @Byte1Value = (@Byte1Value & (POWER(2,3) - 1)) * POWER(2,18),
                  @URL = STUFF(@URL, @Position, 3, ''),
                  @Position = PATINDEX(@Pattern, @URL)
           IF @Position > 0
              SELECT @High = ASCII(UPPER(SUBSTRING(@URL, @Position + 1, 1))) - 48,
                     @Low  = ASCII(UPPER(SUBSTRING(@URL, @Position + 2, 1))) - 48,
                     @High = @High / 17 * 10 + @High % 17,
                     @Low  = @Low  / 17 * 10 + @Low  % 17,
                     @Byte1Value = @Byte1Value + ((16 * @High + @Low) & (POWER(2,6) - 1)) * POWER(2,12),
                     @URL = STUFF(@URL, @Position, 3, ''),
                     @Position = PATINDEX(@Pattern, @URL)
           IF @Position > 0
              SELECT @High = ASCII(UPPER(SUBSTRING(@URL, @Position + 1, 1))) - 48,
                     @Low  = ASCII(UPPER(SUBSTRING(@URL, @Position + 2, 1))) - 48,
                     @High = @High / 17 * 10 + @High % 17,
                     @Low  = @Low  / 17 * 10 + @Low  % 17,
                     @Byte1Value = @Byte1Value + ((16 * @High + @Low) & (POWER(2,6) - 1)) * POWER(2,6),
                     @URL = STUFF(@URL, @Position, 3, ''),
                     @Position = PATINDEX(@Pattern, @URL)
           IF @Position > 0
           BEGIN
              SELECT @High = ASCII(UPPER(SUBSTRING(@URL, @Position + 1, 1))) - 48,
                     @Low  = ASCII(UPPER(SUBSTRING(@URL, @Position + 2, 1))) - 48,
                     @High = @High / 17 * 10 + @High % 17,
                     @Low  = @Low  / 17 * 10 + @Low  % 17,
                     @Byte1Value = @Byte1Value + ((16 * @High + @Low) & (POWER(2,6) - 1))
                     --,@URL = STUFF(@URL, @Position, 3, cast(@Byte1Value as varchar))
                     --,@Position = PATINDEX(@Pattern, @URL)

              SELECT @SurrogateHign = ((@Byte1Value - POWER(16,4)) & (POWER(2,20) - 1)) / POWER(2,10) + 13 * POWER(16,3) + 8 * POWER(16,2),
                     @SurrogateLow = ((@Byte1Value - POWER(16,4)) & (POWER(2,10) - 1)) + 13 * POWER(16,3) + 12 * POWER(16,2),
                     @URL = STUFF(@URL, @Position, 3, NCHAR(@SurrogateHign) + NCHAR(@SurrogateLow)),
                     @Position = PATINDEX(@Pattern, @URL)
           END
       END
    END
    RETURN REPLACE(@URL, '+', ' ') END
于 2019-09-05T22:35:57.563 回答
2
create function [dbo].[URLDecode](
    @str nvarchar( max )
) returns nvarchar( max )
begin
    if @str is null return null
    declare @out nvarchar( max ) = N''
    set @str = replace( @str, N'+', N'%20' )
    while len(@str) > 0
    begin
        declare @i bigint = patindex( N'%[%][0-9a-fA-F][0-9a-fA-F]%', @str )
        if @i = 0 break
        set @out = @out + substring( @str, 0, @i ) + convert( nchar(2), convert( varbinary, '0x' + substring( @str, @i + 1, 2 ), 1 ) )
        set @str = substring( @str, @i + 3, len(@str) )
    end
    return @out + @str
end
于 2017-05-18T02:02:03.660 回答
0

我会添加一些额外的解码。我遇到的一个错误是返回值为空。我还注意到上面的一些答案用空格替换了“+”。

ALTER FUNCTION UrlDecode(@url varchar(3072))
    RETURNS varchar(3072)
AS BEGIN 

    DECLARE @count int, 
            @c char(1), 
            @cenc char(2), 
            @i int, 
            @urlReturn varchar(3072) 

    SET @count = Len(@url) 
    SET @i = 1 
    SET @urlReturn = '' 

    WHILE (@i <= @count) BEGIN 
        SET @c = substring(@url, @i, 1) 
        IF @c LIKE '[!%]' ESCAPE '!' BEGIN 
            SET @cenc = substring(@url, @i + 1, 2) 
            SET @c = CHAR(CASE WHEN SUBSTRING(@cenc, 1, 1) LIKE '[0-9]' 
                            THEN CAST(SUBSTRING(@cenc, 1, 1) as int) 
                            ELSE CAST(ASCII(UPPER(SUBSTRING(@cenc, 1, 1)))-55 as int) 
                            END * 16 + 
                          CASE WHEN SUBSTRING(@cenc, 2, 1) LIKE '[0-9]' 
                            THEN CAST(SUBSTRING(@cenc, 2, 1) as int) 
                            ELSE CAST(ASCII(UPPER(SUBSTRING(@cenc, 2, 1)))-55 as int) 
                            END) 
            SET @urlReturn = @urlReturn + @c 
            SET @i = @i + 2 
        END ELSE BEGIN 
            SET @urlReturn = @urlReturn + @c 
        END 

    SET @i = @i +1 

    END 

    IF @urlReturn is null BEGIN
        set @urlReturn = ''
    END ELSE BEGIN
        set @urlReturn = REPLACE(@urlReturn, '+', ' ')
    END

    RETURN @urlReturn

END
GO
于 2019-10-07T17:15:46.407 回答