SQL Reloaded: Functions Useful in Data Migrations

Left and Right-Padding

Oracle has the LPAD and RPAD functions which return and expression, left-padded, respectively right-padded with the specified characters. The functions are useful in formatting unique identifiers to fit a certain format (eg. 0000012345 instead of 12345). Here’re similar implementations of the respective functions:

-- left padding function 
CREATE FUNCTION dbo.LeftPadding( 
  @str varchar(50) 
, @length int  
, @padchar varchar(1))
RETURNS varchar(50)
AS 
BEGIN 
  RETURN CASE  
    WHEN LEN(@str)<@length THEN Replicate(@padchar, @length-LEN(@str)) + @str       
    ELSE @str  
    END 
END  

-- example left padding 
SELECT dbo.LeftPadding('12345', '10', '0')

-- right padding function 
CREATE FUNCTION dbo.RightPadding( 
  @str varchar(50) 
, @length int  
, @padchar varchar(1))
RETURNS varchar(50)
AS 
BEGIN 
  RETURN CASE  
    WHEN LEN(@str)<@length THEN @str + Replicate(@padchar, @length-LEN(@str))      
	ELSE @str  
  END
END  

-- example right padding 
SELECT dbo.RightPadding('12345', '10', '0')

Left and Right Side

When multiple pieces of data are stored within same same attribute, it’s helpful to get the left, respectively the right part of the string based on a given delimiter, where the reverse flag tells the directions in which the left is applied:

-- left part function 
CREATE FUNCTION dbo.CutLeft( 
  @str varchar(max) 
, @delimiter varchar(1)
, @reverse bit = 0)
RETURNS varchar(max)
AS 
BEGIN 
  RETURN CASE  
     WHEN CharIndex(@delimiter, @str)>0 THEN 
       CASE @reverse 
         WHEN 0 THEN Left(@str, CharIndex(@delimiter, @str)-1)    
         ELSE Left(@str, Len(@str)-CharIndex(@delimiter, Reverse(@str))) 
       END
	 ELSE @str  
  END
END  

-- example left part 
SELECT dbo.CutLeft('12345,045,000', ',', 0)
, dbo.CutLeft('12345,045,000', ',', 1)



-- right part function 
CREATE FUNCTION dbo.CutRight( 
  @str varchar(max) 
, @delimiter varchar(1)
, @reverse bit = 0)
RETURNS varchar(max)
AS 
BEGIN 
  RETURN CASE  
    WHEN CharIndex(@delimiter, @str)>0 THEN 
      CASE @reverse 
        WHEN 0 THEN Right(@str, CharIndex(@delimiter, Reverse(@str))-1) 
        ELSE Right(@str, Len(@str)-CharIndex(@delimiter, @str)) 
      END
   ELSE @str  
  END
END  

-- example right part 
SELECT dbo.CutRight('12345,045,000', ',', 0)
, dbo.CutRight('12345,045,000', ',', 1)

Replacing Special Characters

Special characters can prove to be undesirable in certain scenarios (e.g. matching values, searching). Until a “Replace from” function will be made available, the solution is to include the replacements in a user-defined function similar with the one below:

DROP FUNCTION IF EXISTS [dbo].[ReplaceSpecialChars]

CREATE FUNCTION [dbo].[ReplaceSpecialChars](
@string nvarchar(max)
, @replacer as nvarchar(1) = '-'
) RETURNS nvarchar(max)
-- replaces special characters with a given character (e.g. an empty string, space)
AS
BEGIN   
  IF CharIndex('*', @string) > 0  
     SET @string = replace(@string, '*', @replacer)	
    
  IF CharIndex('#', @string) > 0  
     SET @string = replace(@string, '#', @replacer)	
    
  IF CharIndex('$', @string) > 0  
     SET @string = replace(@string, '$', @replacer)	
    
  IF CharIndex('%', @string) > 0  
     SET @string = replace(@string, '%', @replacer)	
    
  IF CharIndex('&', @string) > 0  
     SET @string = replace(@string, '&', @replacer)	
    
  IF CharIndex(';', @string) > 0  
     SET @string = replace(@string, ';', @replacer)	
    
  IF CharIndex('/', @string) > 0  
     SET @string = replace(@string, '/', @replacer)	
    
  IF CharIndex('?', @string) > 0  
     SET @string = replace(@string, '?', @replacer)	
    
  IF CharIndex('\', @string) > 0  
     SET @string = replace(@string, '\', @replacer)	
    
  IF CharIndex('(', @string) > 0  
     SET @string = replace(@string, '(', @replacer)	
    
  IF CharIndex(')', @string) > 0  
     SET @string = replace(@string, ')', @replacer)	
    
  IF CharIndex('|', @string) > 0  
     SET @string = replace(@string, '|', @replacer)	
    
  IF CharIndex('{', @string) > 0  
     SET @string = replace(@string, '{', @replacer)	
    
  IF CharIndex('}', @string) > 0  
     SET @string = replace(@string, '}', @replacer)	
    
  IF CharIndex('[', @string) > 0  
     SET @string = replace(@string, '[', @replacer)	
    
  IF CharIndex(']', @string) > 0  
     SET @string = replace(@string, ']', @replacer)	
                                
  RETURN (LTrim(RTrim(@string)))
END

SELECT [dbo].[ReplaceSpecialChars]('1*2#3$4%5&6;7/8?9\10(11)12|13{14}15[16]', '')
SELECT [dbo].[ReplaceSpecialChars]('1*2#3$4%5&6;7/8?9\10(11)12|13{14}15[16]', ' ')

Other type of special characters are the umlauts (e.g. ä, ß, ö, ü from German language):

DROP FUNCTION IF EXISTS [dbo].[ReplaceUmlauts]

CREATE FUNCTION [dbo].[ReplaceUmlauts](
@string nvarchar(max)
) RETURNS nvarchar(max)
-- replaces umlauts with their equivalent
AS
BEGIN   
  IF CharIndex('ä', @string) > 0  
     SET @string = replace(@string, 'ä', 'ae')	
    
  IF CharIndex('ö', @string) > 0  
     SET @string = replace(@string, 'ö', 'oe')	
    
  IF CharIndex('ß', @string) > 0  
     SET @string = replace(@string, 'ß', 'ss')	
    
  IF CharIndex('%', @string) > 0  
     SET @string = replace(@string, 'ü', 'ue')	
    
                                
  RETURN Trim(@string)
END

SELECT [dbo].[ReplaceUmlauts]('Hr Schrötter trinkt ein heißes Getränk')

Handling Umlauts

Another type of specials characters are the letter specific to certain languages that deviate from the Latin characters (e.g. umlauts in German, accents in French), the usage of such characters introducing further challenges in handling the characters, especially when converting the data between characters sets. A common scenario is the one in which umlauts in ISO-8891-1 are encoded using two character sets. Probably the easiest way to handle such characters is to write a function as follows:

-- create the function 
CREATE FUNCTION [dbo].[ReplaceCodes2Umlauts](
  @string nvarchar(max)
) RETURNS nvarchar(max)
-- replaces ISO 8859-1 characters with the corresponding encoding 
AS
BEGIN   
  IF CharIndex('ß', @string) > 0  
     SET @string = replace(@string, 'ß', 'ß')	

  IF CharIndex('ö', @string) > 0  
     SET @string = replace(@string, 'ö', 'ö')	

  IF CharIndex('Ö', @string) > 0  
     SET @string = replace(@string, 'Ö', 'Ö')	
    
  IF CharIndex('ü', @string) > 0  
     SET @string = replace(@string, 'ü', 'ü')	

  IF CharIndex('Ü', @string) > 0  
     SET @string = replace(@string, 'Ãœ', 'Ü')	

  IF CharIndex('ä', @string) > 0  
     SET @string = replace(@string, 'ä', 'ä')

  IF CharIndex('Ä', @string) > 0  
     SET @string = replace(@string, 'Ä', 'Ä')
                                
  RETURN (LTrim(RTrim(@string)))
END


--test the function 
SELECT [dbo].[ReplaceCodes2Umlauts]('Falsches Ãœben von Xylophonmusik quält jeden größeren Zwerg')

In the inverse scenario, at least for the German language is possible to replace the umlauts with a set the corresponding transliterations:

-- drop the function 
DROP FUNCTION IF EXISTS [dbo].[ReplaceUmlauts]

-- create the function 
CREATE FUNCTION [dbo].[ReplaceUmlauts](
  @string nvarchar(max)
) RETURNS nvarchar(max)
-- replaces umlauts with corresponding transliterations 
AS
BEGIN   
  IF CharIndex('ß', @string) > 0  
     SET @string = replace(@string, 'ß', 'ss')	

  IF CharIndex('ü', @string) > 0  
     SET @string = replace(@string, 'ü', 'ue')	

  IF CharIndex('ä', @string) > 0  
     SET @string = replace(@string, 'ä', 'ae')	
    
  IF CharIndex('ö', @string) > 0  
     SET @string = replace(@string, 'ö', 'oe')	
                                
  RETURN (LTrim(RTrim(@string)))
END

--test the function 
SELECT [dbo].[ReplaceUmlauts]('Falsches üben von Xylophonmusik quält jeden größeren Zwerg')

Happy coding!