replace('{here it is' , '{', '^') doesn't work

i'm trying to do a replacement function and it works for everything but { characters
so i thought i would fake it out and replace the { and } with ^
but i get an incorrect syntax
where @originalcode is the string
Set @OriginalCode = Replace(@OriginalCode, '{', '^^')
Set @OriginalCode = Replace(@OriginalCode, '}', '^')
quigleyryanAsked:
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

BrandonGalderisiCommented:
Not sure... Because this works:
declare @originalcode nvarchar(max)
set @originalcode = '{here i}t is'
set @originalcode = replace(@originalcode , '{','^^')
set @originalcode = replace(@originalcode , '}','^')
select @originalcode

Open in new window

0
BrandonGalderisiCommented:
Ultimately, what is your goal?  Is { some soft of a delimiter?
0
quigleyryanAuthor Commented:
its a parsing routine - i have a lot of "junk" and print specific characters/instructions in my data - i need to replace the print specific stuff with html code...
so initially i had a function (which works with everything but { brackets) that goes and does the replaces based on a set of compares
so if the data had as @originalcode = '{\this way now }' my code would make it <sup>This way now </sup>
0
Ultimate Tool Kit for Technology Solution Provider

Broken down into practical pointers and step-by-step instructions, the IT Service Excellence Tool Kit delivers expert advice for technology solution providers. Get your free copy now.

BrandonGalderisiCommented:
I think the problem exists in that function then.  With only seeing a small section of the function, it's hard to say.  All I know is a search and replace on { and } are valid.
0
quigleyryanAuthor Commented:
Here's the whole thing...

well this is what i would rather do --
i have this scalar-valued function that is called by a stored procedure with
this command: SET @OriginalCode = dbo.ReplaceTokenPair(@OriginalCode, '{b',
'}', '<b>', '</b>')

If i have @originalcode = 'testing {b this is a test }' which should give me
testing <b> this is a test </b> but instead it goes into never never land


ALTER FUNCTION [dbo].[ReplaceTokenPair]
(
      @OriginalCode nvarchar(4000),
      @StartToken nvarchar(100),
      @EndToken nvarchar(100),
      @StartHTML nvarchar(100),
      @EndHTML nvarchar(100)
)
RETURNS nvarchar(4000)
AS
BEGIN
      DECLARE @Before nvarchar(4000), @Between nvarchar(4000), @After
nvarchar(4000)

      DECLARE @x int, @y int, @z int
      SET @x = 1
      SET @y = 1
      SET @z = 1

      WHILE @y > 0 AND @z > 0
      BEGIN
            SET @y = CHARINDEX(@StartToken, @OriginalCode, @x)
            SET @z = CHARINDEX(@EndToken, @OriginalCode, @y + LEN(@StartToken) + 1)

            IF @y > 0
            BEGIN
                  SET @Before = SUBSTRING(@OriginalCode, 1, @y - 1)
                  SET @Between = SUBSTRING(@OriginalCode, @y + LEN(@StartToken), @z - @y -
LEN(@EndToken) - 1)
                  SET @After = SUBSTRING(@OriginalCode, @z + LEN(@EndToken),
LEN(@OriginalCode))
                  SET @OriginalCode = @Before + @StartHTML + @Between + @EndHTML + @After
            END
            SET @x = LEN(@OriginalCode) - (LEN(@Before) + LEN(@StartHTML) +
LEN(@Between) + LEN(@EndHTML))
            SET @x = 0
      END
0
BrandonGalderisiCommented:
Try this:



alter FUNCTION [dbo].[ReplaceTokenPair]
(
      @OriginalCode nvarchar(max),
      @StartToken nvarchar(100),
      @EndToken nvarchar(100),
      @StartHTML nvarchar(100),
      @EndHTML nvarchar(100)
)
RETURNS nvarchar(max)
AS
BEGIN
declare @Return nvarchar(max)
select @Return = isnull(@Return,'')+  @StartHTML + substring(theValue, 1,charindex(@EndToken,theValue)-1) + @EndHTML
+ substring(theValue,charindex(@endToken,theValue)+1,(datalength(theValue)/2)-charindex(@endToken,theValue)+1)
from [dbo].[fn_DelimitedToTable](@OriginalCode, @StartToken)
where datalength(theValue)/2>0

return @Return
end

go
select dbo.ReplaceTokenPair('{bhello}', '{b','}', '<b>', '</b>')



It depends upon this function:
if object_id('[dbo].[fn_DelimitedToTable]') is not null
     drop function [dbo].[fn_DelimitedToTable]
go
create function [dbo].[fn_DelimitedToTable](@DelimitedString nvarchar(max), @Delimiter nvarchar(32))
returns @Values TABLE
     (ident         int not null identity primary key clustered
     ,thePosition   int not null
     ,theValue      nvarchar(max)
     )
as
/************************************************************
*
*    Author:        Brandon Galderisi
*    Last modified: 07-Oct-2008
*    Purpose:       splits an input string (@DelimitedString) 
*                   on a delimiter (@delimiter) and outputs 
*                   a table of values.
*    
*
*************************************************************/
begin
 
insert into @Values (thePosition,theValue)
		select n, substring(@delimiter + @DelimitedString + @delimiter, n + (datalength(@delimiter)/2), charindex(@delimiter, @delimiter + @DelimitedString + @delimiter, n + datalength(@delimiter)/2) - n - datalength(@delimiter)/2) as string_value
		from	dbo.vw_Nums
		where
			n <= (datalength(@delimiter + @DelimitedString + @delimiter)/2) - (datalength(@delimiter)/2)
			and substring(@delimiter + @DelimitedString + @delimiter, n, (datalength(@delimiter)/2)) = @delimiter
 
 
 
return
end
/*
-- The purpose of vw_Nums is if the source database does not have a numbers table.  This view (vw_nums)
-- will allow the parsing of up to 4 billion+ character strings with the above function.  Whether a static
-- table or this view is used for fn_DelimitedToTable, it can only split a string as long as the source 
-- numbers table.
Requires a nunbers table or this view:
create view vw_Nums
as
with   cte0 as (select 1 as c union all select 1), 
       cte1 as (select 1 as c from cte0 a, cte0 b), 
       cte2 as (select 1 as c from cte1 a, cte1 b), 
       cte3 as (select 1 as c from cte2 a, cte2 b), 
       cte4 as (select 1 as c from cte3 a, cte3 b), 
       cte5 as (select 1 as c from cte4 a, cte4 b), 
       nums as (select row_number() over (order by c) as n from cte5)
       select n from nums 
 
 
 
Sample Usage:
select * from [dbo].[fn_DelimitedToTable]('a|%25basdf|%25c|%25d','|%25')
select theValue from [dbo].[fn_DelimitedToTable]('a','|')
select * from [dbo].[fn_DelimitedToTable]('a basdf c d',' ')
*/
GO

Open in new window

0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
BrandonGalderisiCommented:
Actually... that one only works for one } right now.... give me a few.
0
BrandonGalderisiCommented:
I have a potential problem to propose to you.  If you use the same terminator for multiple tags, there is no reliable way to parse it.

So if you have:
{b{ihello}my name is} jon

There is nothing to say that the first } goes with {b or {i.  My process that I'm working on now will find all terminators '}' and not just the first one.  So it would also not work well with

{bhello}my name is} jon

you would get
<b>hello</b>my name is</b> jon
0
quigleyryanAuthor Commented:
Brandon
thank you - iright now the data has print specific characters in it - and luckily as we go through and parse the strings we don't have a scenario of {b{ihello}my name is}
but i guess i could...
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Query Syntax

From novice to tech pro — start learning today.

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.