CODEWITHSUNDEEP
sqlsql-serversql-server-2008t-sqlreplace
Saharsh Shah
Saharsh Shah

Reputation: 29051

Replace multiple characters from string without using any nested replace functions

I have an equation stored in my table. I am fetching one equation at a time and want to replace all the operators with any other character.

Input String: (N_100-(6858)*(6858)*N_100/0_2)%N_35

Operators or patterns: (+, -, *, /, %, (, ))

Replacement character: ~

Output String: ~N_100~~6858~~~6858~~N_100~0_2~~N_35

I had tried below query with Nested REPLACE Functions and I got desired output: 

DECLARE @NEWSTRING VARCHAR(100) 
SET @NEWSTRING = '(N_100-(6858)*(6858)*N_100/0_2)%N_35' ;
SELECT @NEWSTRING = REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(
                    @NEWSTRING, '+', '~'), '-', '~'), '*', '~'), '/', '~')
                   , '%', '~'), '(', '~'), ')', '~')
PRINT @NEWSTRING

Output: ~N_100~~6858~~~6858~~N_100~0_2~~N_35

How can I replace all the operators without using nested replace functions?

Upvotes: 6

Views: 47223

Answers (5)

komal siddhapura
komal siddhapura

Reputation: 1

you can replace all the operators without using nested replace functions

DECLARE @Pattern VARCHAR(300) ='%[+-\*/%()]%';
DECLARE @String VARCHAR(300) ='(N_100-(6858)*(6858)*N_100/0_2)%N_35';
WHILE PatIndex(@Pattern, @String) <> 0
    SELECT @String=Replace(@String, Substring(@String, PatIndex(@Pattern, @String), 1), '~')
SELECT @String

Upvotes: 0

Lukasz Szozda
Lukasz Szozda

Reputation: 175736

The easiest way is to use TRANSLATE function. It is availble from SQL Server 2017 (aka vNext) and above.

TRANSLATE

Returns the string provided as a first argument after some characters specified in the second argument are translated into a destination set of characters.

TRANSLATE ( inputString, characters, translations)

Returns a character expression of the same type as inputString where characters from the second argument are replaced with the matching characters from third argument.

In your case:

SELECT TRANSLATE('(N_100-(6858)*(6858)*N_100/0_2)%N_35', '+-*/%()','~~~~~~~')

DBFiddle Demo

Upvotes: 2

Shane Gebs
Shane Gebs

Reputation: 121

I believe it is easier and more readable if you use a table to drive this. 

declare @String varchar(max) = '(N_100-(6858)*(6858)*N_100/0_2)%N_35'

--table containing values to be replaced
create table #Replace 
(
    StringToReplace varchar(100) not null primary key clustered
    ,ReplacementString varchar(100) not null    
)

insert into #Replace (StringToReplace, ReplacementString)
values ('+', '~')
    ,('-', '~')
    ,('*', '~')
    ,('/', '~')
    ,('%', '~')
    ,('(', '~')
    ,(')', '~')

select @String = replace(@String, StringToReplace, ReplacementString)
from #Replace a

select @String

drop table #Replace

Upvotes: 12

Saharsh Shah
Saharsh Shah

Reputation: 29051

I had created a SPLIT function to implement this because I need to implement this operation multiple time in PROCEDURE

SPLIT FUNCTION

create function [dbo].[Split](@String varchar(8000), @Delimiter char(1))       
returns @temptable TABLE (items varchar(8000))       
as       
begin       
    declare @idx int       
    declare @slice varchar(8000)       

    select @idx = 1       
        if len(@String)<1 or @String is null  return       

    while @idx!= 0       
    begin       
        set @idx = charindex(@Delimiter,@String)       
        if @idx!=0       
            set @slice = left(@String,@idx - 1)       
        else       
            set @slice = @String       

        if(len(@slice)>0)  
            insert into @temptable(Items) values(@slice)       

        set @String = right(@String,len(@String) - @idx)       
        if len(@String) = 0 break       
    end   
return       
end

Code used in procedure: 

DECLARE @NEWSTRING VARCHAR(100) 
SET @NEWSTRING = '(N_100-(6858)*(6858)*N_100/0_2)%N_35' ;
SELECT @NEWSTRING = REPLACE(@NEWSTRING, items, '~') FROM dbo.Split('+,-,*,/,%,(,)', ',');
PRINT @NEWSTRING

OUTPUT

~N_100~~6858~~~6858~~N_100~0_2~~N_35

Upvotes: 0

Twinkles
Twinkles

Reputation: 1994

There is not equivalent for the TRANSLATE function from Oracle in SQL Server, you have to use nested replace functions.

The following solution is technically correct:

DECLARE @newstring VARCHAR(100) = '(N_100-(6858)*(6858)*N_100/0_2)%N_35';
DECLARE @pattern VARCHAR(100) = '%[+-\*/%()]%';
DECLARE @i INT;
BEGIN
  SET @i = PATINDEX(@pattern,@newstring) 
  WHILE @i <> 0
  BEGIN
    SET @newstring = LEFT(@newstring,@i-1) + '~' + SUBSTRING(@newstring,@i+1,100);
    SET @i = PATINDEX(@pattern,@newstring) 
  END
  SELECT @newstring;
END;

But I do not see why you would favor this over nested REPLACE calls.

Upvotes: 5

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