Question

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我正在尝试用URL替换字符串中的特殊字符编码它们对应的值。下面是我的一些示例代码一直在努力。

感谢您的帮助。

create table #url_encoding_lookup(character varchar(10), code varchar (20))

insert into  #url_encoding_lookup (character, code)
values 
('!',   '%21'),
('"',   '%22'),
('#',   '%23'),
('$',   '%24'),
('%',   '%25'),
('&',   '%26'),
('''',  '%27'),
('(',   '%28'),
(')',   '%29'),
('*',   '%2A'),
('+',   '%2B'),
(',',   '%2C'),
('-',   '%2D'),
('.',   '%2E'),
('/',   '%2F')


Create   table #data
(string varchar (200))

insert into #data
values

('Jim (BoB)'),
('Will''s Place'),
('Auto-Mart')


select * from #data
select * from #url_encoding_lookup

desired results would be 

Jim %28Bob%29
Will%27s Place
Auto%2DMart

Answer 1

ALTER FUNCTION [dbo].[udf_ReplaceYouCoded]
(
@the_string nvarchar(max) 
)
RETURNS NVARCHAR(MAX)
AS
BEGIN
declare @temp_field nvarchar(max)

create table #url_encoding_lookup(character varchar(10), code varchar (20))

insert into  #url_encoding_lookup (character, code)
values 
('!',   '%21'),
('"',   '%22'),
('#',   '%23'),
('$',   '%24'),
('%',   '%25'),
('&',   '%26'),
('''',  '%27'),
('(',   '%28'),
(')',   '%29'),
('*',   '%2A'),
('+',   '%2B'),
(',',   '%2C'),
('-',   '%2D'),
('.',   '%2E'),
('/',   '%2F')

declare @x as int
set @x = 1

--LOOP @the_string 
while @i < len(@the_string)

if(substring(@the_string,x,1) = (select character from #url_encoding_lookup where character = substring(@the_string,x,1)))
begin
    @temp_field = @temp_field + (select code from #url_encoding_lookup where character = substring(@the_string,x,1))
end
else
begin
    @temp_field = @temp_field + substring(@the_string,x,1)
end

select @x = @x + 1

end

RETURN @temp_field 
END

Answer 2

创建程序

BEGIN 
  DECLARE _end BOOLEAN DEFAULT FALSE;
  DECLARE _result CHAR(200) DEFAULT str; 
  DECLARE _find VARCHAR(32);
  DECLARE _replace VARCHAR(32);
  DECLARE _cur CURSOR FOR SELECT _character, _code FROM url_encoding_lookup; 
  DECLARE CONTINUE HANDLER FOR NOT FOUND SET _end = TRUE; 

  OPEN _cur; 
    _loop: LOOP 
      FETCH _cur INTO _find, _replace; 
        IF _end THEN 
          LEAVE _loop; 
        END IF; 
       SET _result = REPLACE(_result, _find, _replace);
     END LOOP _loop; 
  CLOSE _cur; 
  RETURN _result; 
END

然后

SELECT _replace_chars(name) FROM `data`

结果

Jim %28BoB%29
Will%27s Place
Auto%2DMart

Answer 3

ALTER function [dbo].[udf_ReplaceSpecialChars]
 (@s varchar(256))

 returns varchar(256)
  as
begin
-- declare @s varchar(256) set @s = 'Jim (P)'  --test
   if @s is null
  return null
   declare @s2 varchar(256)
   set @s2 = '' --set variable to empty string. ready to recieve values
  declare @l int
  set @l = len(@s) --determin the number of characters in @s
   declare @p int
   set @p = 1  --set beginning string position
   while @p <= @l begin
   declare @c int
   set @c = ascii(substring(@s, @p, 1)) --find the ascii number for 1st 
character
  declare @nc varchar(256)
  set @nc = (select code from url_encoding_lookup where ascii_code = @c) --
get corresponding URL encoding string from lookup table
   if @c between 33 and 47 or @c between 58 and 64 or @c between 91 and 96 
or @c between 123 and 255 --when looping through each character, if special 
character 
     set @s2 = @s2 +  @nc --then use string from lookup table
   else if  @c = 32 or @c between 48 and 57 or @c between 65 and 90 or @c 
between 97 and 122 --if character is not special
     set @s2 = @s2 + char(@c) --then find char value of character

  set @p = @p + 1 --set position to next charachter for loop to look at
  end   
   if len(@s2) = 0
  return null
   return @s2 --return rebuilt string
  end

Answer 4

您可以使用以下正则表达式[\u0100-\uffff]替换特殊字符

select regexp_replace(column, '[\u0100-\uffff]', '')

SQL使用另一个表中的值替换特殊字符

4 个答案: