Using RegEx in SQL Server

Question

I'm looking how to replace/encode text using RegEx based on RegEx settings/params below:

RegEx.IgnoreCase = True     
RegEx.Global = True     
RegEx.Pattern = "[^a-z\d\s.]+"   

I have seen some examples on RegEx, but confused as to how to apply it the same way in SQL Server. Any suggestions would be helpful. Thank you.

Answer 1

You do not need to interact with managed code, as you can use LIKE:

CREATE TABLE #Sample(Field varchar(50), Result varchar(50))
GO
INSERT INTO #Sample (Field, Result) VALUES ('ABC123 ', 'Do not match')
INSERT INTO #Sample (Field, Result) VALUES ('ABC123.', 'Do not match')
INSERT INTO #Sample (Field, Result) VALUES ('ABC123&', 'Match')
SELECT * FROM #Sample WHERE Field LIKE '%[^a-z0-9 .]%'
GO
DROP TABLE #Sample

As your expression ends with + you can go with '%[^a-z0-9 .][^a-z0-9 .]%'

EDIT:
To make it clear: SQL Server doesn't support regular expressions without managed code. Depending on the situation, the LIKE operator can be an option, but it lacks the flexibility that regular expressions provides.

Answer 2

Regular Expressions In SQL Server Databases Implementation Use

Regular Expression - Description
. Match any one character
* Match any character
+ Match at least one instance of the expression before
^ Start at beginning of line
$ Search at end of line
< Match only if word starts at this point
> Match only if word stops at this point
\n Match a line break
[] Match any character within the brackets
[^...] Matches any character not listed after the ^
[ABQ]% The string must begin with either the letters A, B, or Q and can be of any length
[AB][CD]% The string must have a length of two or more and which must begin with A or B and have C or D as the second character
[A-Z]% The string can be of any length and must begin with any letter from A to Z
[A-Z0-9]% The string can be of any length and must start with any letter from A to Z or numeral from 0 to 9
[^A-C]% The string can be of any length but cannot begin with the letters A to C
%[A-Z] The string can be of any length and must end with any of the letters from A to Z
%[%$#@]% The string can be of any length and must contain at least one of the special characters enclosed in the bracket

Answer 3

You will have to build a CLR procedure that provides regex functionality, as this article illustrates.

Their example function uses VB.NET:

Imports System
Imports System.Data.Sql
Imports Microsoft.SqlServer.Server
Imports System.Data.SqlTypes
Imports System.Runtime.InteropServices
Imports System.Text.RegularExpressions
Imports System.Collections 'the IEnumerable interface is here  


Namespace SimpleTalk.Phil.Factor
    Public Class RegularExpressionFunctions
        'RegExIsMatch function
        <SqlFunction(IsDeterministic:=True, IsPrecise:=True)> _
        Public Shared Function RegExIsMatch( _
                                            ByVal pattern As SqlString, _
                                            ByVal input As SqlString, _
                                            ByVal Options As SqlInt32) As SqlBoolean
            If (input.IsNull OrElse pattern.IsNull) Then
                Return SqlBoolean.False
            End If
            Dim RegExOption As New System.Text.RegularExpressions.RegExOptions
            RegExOption = Options
            Return RegEx.IsMatch(input.Value, pattern.Value, RegExOption)
        End Function
    End Class      ' 
End Namespace

...and is installed in SQL Server using the following SQL (replacing '%'-delimted variables by their actual equivalents:

sp_configure 'clr enabled', 1
RECONFIGURE WITH OVERRIDE

IF EXISTS ( SELECT   1
            FROM     sys.objects
            WHERE    object_id = OBJECT_ID(N'dbo.RegExIsMatch') ) 
   DROP FUNCTION dbo.RegExIsMatch
go

IF EXISTS ( SELECT   1
            FROM     sys.assemblies asms
            WHERE    asms.name = N'RegExFunction ' ) 
   DROP ASSEMBLY [RegExFunction]

CREATE ASSEMBLY RegExFunction 
           FROM '%FILE%'
GO

CREATE FUNCTION RegExIsMatch
   (
    @Pattern NVARCHAR(4000),
    @Input NVARCHAR(MAX),
    @Options int
   )
RETURNS BIT
AS EXTERNAL NAME 
   RegExFunction.[SimpleTalk.Phil.Factor.RegularExpressionFunctions].RegExIsMatch
GO

--a few tests
---Is this card a valid credit card?
SELECT dbo.RegExIsMatch ('^(?:4[0-9]{12}(?:[0-9]{3})?|5[1-5][0-9]{14}|6(?:011|5[0-9][0-9])[0-9]{12}|3[47][0-9]{13}|3(?:0[0-5]|[68][0-9])[0-9]{11}|(?:2131|1800|35\d{3})\d{11})$','4241825283987487',1)
--is there a number in this string
SELECT dbo.RegExIsMatch( '\d','there is 1 thing I hate',1)
--Verifies number Returns 1
DECLARE @pattern VARCHAR(255)
SELECT @pattern ='[a-zA-Z0-9]\d{2}[a-zA-Z0-9](-\d{3}){2}[A-Za-z0-9]'
SELECT  dbo.RegExIsMatch (@pattern, '1298-673-4192',1),
        dbo.RegExIsMatch (@pattern,'A08Z-931-468A',1),
        dbo.RegExIsMatch (@pattern,'[A90-123-129X',1),
        dbo.RegExIsMatch (@pattern,'12345-KKA-1230',1),
        dbo.RegExIsMatch (@pattern,'0919-2893-1256',1)

Answer 4

Slightly modified version of Julio's answer.

-- MS SQL using VBScript Regex
-- select dbo.RegexReplace('aa bb cc','($1) ($2) ($3)','([^\s]*)\s*([^\s]*)\s*([^\s]*)')
-- $$ dollar sign, $1 - $9 back references, $& whole match

CREATE FUNCTION [dbo].[RegexReplace]
(   -- these match exactly the parameters of RegExp
    @searchstring varchar(4000),
    @replacestring varchar(4000),
    @pattern varchar(4000)
)
RETURNS varchar(4000)
AS
BEGIN
    declare @objRegexExp int, 
        @objErrorObj int,
        @strErrorMessage varchar(255),
        @res int,
        @result varchar(4000)

    if( @searchstring is null or len(ltrim(rtrim(@searchstring))) = 0) return null
    set @result=''
    exec @res=sp_OACreate 'VBScript.RegExp', @objRegexExp out
    if( @res <> 0) return '..VBScript did not initialize'
    exec @res=sp_OASetProperty @objRegexExp, 'Pattern', @pattern
    if( @res <> 0) return '..Pattern property set failed'
    exec @res=sp_OASetProperty @objRegexExp, 'IgnoreCase', 0
    if( @res <> 0) return '..IgnoreCase option failed'
    exec @res=sp_OAMethod @objRegexExp, 'Replace', @result OUT,
         @searchstring, @replacestring
    if( @res <> 0) return '..Bad search string'
    exec @res=sp_OADestroy @objRegexExp
    return @result
END

You'll need Ole Automation Procedures turned on in SQL:

exec sp_configure 'show advanced options',1; 
go
reconfigure; 
go
sp_configure 'Ole Automation Procedures', 1; 
go
reconfigure; 
go
sp_configure 'show advanced options',0; 
go
reconfigure;
go

Answer 5

SELECT * from SOME_TABLE where NAME like '%[^A-Z]%'

Or some other expression instead of A-Z

Answer 6

A similar approach to @mwigdahl's answer, you can also implement a .NET CLR in C#, with code such as;

using System.Data.SqlTypes;
using RX = System.Text.RegularExpressions;

public partial class UserDefinedFunctions
{
 [Microsoft.SqlServer.Server.SqlFunction]
 public static SqlString Regex(string input, string regex)
 {
  var match = RX.Regex.Match(input, regex).Groups[1].Value;
  return new SqlString (match);
 }
}

Installation instructions can be found here