String User-Defined Functions

Introduction

String UDFs

StrIns

StrDel

StrSeparate

StrCHARINDEX

StrREPLACE

StrREVERSE

Introduction

I would like to write the series of articles about useful User-Defined
Functions grouped by the following categories:

Date and Time User-Defined Functions

Mathematical User-Defined Functions

Metadata User-Defined Functions

Security User-Defined Functions

String User-Defined Functions

System User-Defined Functions

Text and Image User-Defined Functions

In this article, I wrote some useful String User-Defined Functions.

String UDFs

These scalar User-Defined Functions perform an operation on a string
input value and return a string or numeric value.

StrIns

Inserts set of characters into another set of characters at a specified
starting point.
Syntax
StrIns ( character_expression, start, character_expression )
Arguments
character_expression
Is an expression of character data. character_expression can be a
constant, variable, or column of character data.
start
Is an integer value that specifies the location to begin insertion.
If start or length is negative, a null string is returned. If start
is longer than the first character_expression, a null string is
returned.
Return Types
nvarchar
The function’s text:




CREATE FUNCTION StrIns
  ( @str_1 nvarchar(4000),
    @start int,
    @str_2 nvarchar(4000) )
RETURNS nvarchar(4000)
AS
BEGIN
  RETURN (STUFF (@str_1, @start, 0, @str_2))
END
GO




Examples
This example returns a character string created by inserting the
second string starting at position 2 (at b) into the first string.




SELECT dbo.StrIns(‘abcdef’, 2, ‘ijklmn’)
GO




Here is the result set:
————
aijklmnbcdef
(1 row(s) affected)

StrDel

Deletes a specified length of characters at a specified starting point.
Syntax
StrDel ( character_expression, start, length )
Arguments
character_expression
Is an expression of character data. character_expression can be a
constant, variable, or column of character data.
start
Is an integer value that specifies the location to begin deletion.
If start or length is negative, a null string is returned. If start
is longer than the first character_expression, a null string is
returned.
length
Is an integer that specifies the number of characters to delete.
If length is longer than the first character_expression, deletion
occurs up to the last character in the last character_expression.
Return Types
nvarchar
The function’s text:




CREATE FUNCTION StrDel
  ( @str_1 nvarchar(4000),
    @start int,
    @length int )
RETURNS nvarchar(4000)
AS
BEGIN
  RETURN (STUFF (@str_1 , @start, @length, ”))
END
GO




Examples
This example returns a character string created by deleting three
characters from the first string (abcdef) starting at position 2
(at b).




SELECT dbo.StrDel(‘abcdef’, 2, 3)
GO




Here is the result set:
—
aef
(1 row(s) affected)

StrSeparate

Inserts a specified character into the given string after
every n-th character (from the end of the string).
Syntax
StrSeparate ( character_expression, term, number )
Arguments
character_expression
Is an expression of character data. character_expression can be a
constant, variable, or column of character data.
term
Is a character.
number
Is an integer.
Return Types
nvarchar
The function’s text:




CREATE FUNCTION StrSeparate
  ( @str nvarchar(4000),
    @term char(1),
    @number int )
RETURNS nvarchar(4000)
AS
BEGIN
  DECLARE @i int, @j int, @stepcount int
  IF (len(@str) <= @number) RETURN @str
  SELECT @str =REVERSE(@str), @i = 1, @j = @number + 1,
         @stepcount = len(@str) / @number
  WHILE @i <= @stepcount
    BEGIN
      SET @str = ISNULL(STUFF(@str, @j, 0, @term), @str)
      SET @j = @j + @number + 1
      SET @i = @i + 1
    END
  SET @str = REVERSE(@str)
  RETURN @str
END
GO




Examples
This example returns a character string created by inserting the
space character after every three characters of the specified
string (from the end of the string).




SELECT dbo.StrSeparate(‘12345678’, ‘ ‘, 3)
GO




Here is the result set:
———-
12 345 678
(1 row(s) affected)

StrCHARINDEX

Returns the starting position of the n-th entering of the specified
expression in a character string.
Syntax
CHARINDEX ( expression1, expression2, start_location, number)
Arguments
expression1
Is an expression containing the sequence of characters to be found.
expression1 is an expression of the short character data type category.
expression2
Is an expression, usually a column searched for the specified sequence.
expression2 is of the character string data type category.
start_location
Is the character position to start searching for expression1 in
expression2. If start_location is a negative number, or is zero,
the search starts at the beginning of expression2.
number
Is an integer.
Return Types
int
The function’s text:




CREATE FUNCTION StrCHARINDEX
  ( @expression1 nvarchar(4000),
    @expression2 nvarchar(4000),
    @start_location int = 0,
    @number int )
RETURNS int
AS
BEGIN
  DECLARE @i int, @position int
  SET @i = 1
  WHILE (@i <= @number) AND (CHARINDEX(@expression1, @expression2, 
@start_location) <> 0)
    BEGIN
      SET @position = CHARINDEX(@expression1, @expression2, @start_location)
      SET @expression2 = STUFF(@expression2,
                               CHARINDEX(@expression1, @expression2,
@start_location),
                               len(@expression1),
                               space(len(@expression1)))
      SET @i = @i + 1
    END
  RETURN @position
END
GO




Examples




SELECT dbo.StrCHARINDEX(’12’, ‘2312451267124’, 0, 2)
GO




Here is the result set:
———–
7
(1 row(s) affected)

StrREPLACE

Replaces all occurrences of the second given string expression in the
first string expression with a third expression starting from the
start_location position.
Syntax
REPLACE(‘string_expression1′,’string_expression2′,’string_expression3’,@start_location)
Arguments
‘string_expression1’
Is the string expression to be searched.
‘string_expression2’
Is the string expression to try to find.
‘string_expression3’
Is the replacement string expression.
start_location
Is the character position to start replacing.
Return Types
nvarchar
The function’s text:




CREATE FUNCTION StrREPLACE
  ( @string_expression1 nvarchar(4000),
    @string_expression2 nvarchar(4000),
    @string_expression3 nvarchar(4000),
    @start_location int )
RETURNS nvarchar(4000)
AS
BEGIN
  IF (@start_location <= 0) OR (@start_location > len(@string_expression1))
    RETURN (REPLACE (@string_expression1, @string_expression2,
@string_expression3))
  RETURN (STUFF (@string_expression1,
                 @start_location,
                 len(@string_expression1) – @start_location + 1,
                 REPLACE(SUBSTRING (@string_expression1,
                                    @start_location,
                                    len(@string_expression1) –
@start_location + 1),
                                    @string_expression2,
                                    @string_expression3)))
END
GO




Examples




SELECT dbo.StrREPLACE(‘12345678912345′, ’23’, ‘**’, 4)
GO




Here is the result set:
——————-
1234567891**45
(1 row(s) affected)

StrREVERSE

Returns the reverse of a character expression starting at
the specified position.
Syntax
REVERSE ( character_expression, start_location )
Arguments
character_expression
Is an expression of character data. character_expression can be
a constant, variable, or column of character data.
start_location
Is the character position to start reversing.
Return Types
nvarchar
The function’s text:




CREATE FUNCTION StrREVERSE
  ( @character_expression  nvarchar(4000),
    @start_location int )
RETURNS nvarchar(4000)
AS
BEGIN
  IF (@start_location <= 0) OR (@start_location >
len(@character_expression))
    RETURN (REVERSE(@character_expression))
  RETURN (STUFF (@character_expression,
                 @start_location,
                 len(@character_expression) – @start_location + 1,
                 REVERSE(SUBSTRING (@character_expression,
                                    @start_location,
                                    len(@character_expression) –
@start_location + 1))))
END
GO




Examples




SELECT dbo.StrREVERSE(‘123456789’, 3)
GO




Here is the result set:
——————-
129876543
(1 row(s) affected)

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