I'm trying to emulate Oracle's RTRIM(expression, characters)
in MsSql Server 2008 R2 with the following query:
REVERSE(
SUBSTRING(
REVERSE(field),
PATINDEX('%[^chars]%', REVERSE(field)),
LEN(field) - PATINDEX('%[^chars]%', REVERSE(field)) + 1
)
)
The problem is that I want to be able to trim characters like ]
and ^
which do probably need escaping.
I don't know how to do this. Things like \]
don't work.
I'm aware of the ESCAPE
clause but I do not understand exactly how it works and, by the way, SqlServer refuses it if put right after the pattern string.
Fun fact:
If I write %[^^]%
(desiring to trim ^
) it doesn't work.
If I write %[^ ^]%
it does trim ^
, but clearly also trim spaces!
Not pretty, but...
CREATE FUNCTION dbo.RTRIMCHARS(
@input AS VARCHAR(MAX), @chars AS VARCHAR(100)
) RETURNS VARCHAR(MAX)
AS
BEGIN
DECLARE @charpos BIGINT
DECLARE @strpos BIGINT
SET @strpos = LEN(@input)
SET @charpos = LEN(@chars)
IF @strpos IS NULL OR @charpos IS NULL RETURN NULL
IF @strpos = 0 OR @charpos = 0 RETURN @input
WHILE @strpos > 0
BEGIN
SET @charpos = LEN(@chars)
WHILE @charpos > 0
BEGIN
IF SUBSTRING(@chars, @charpos, 1) = SUBSTRING(@input, @strpos, 1)
BEGIN
SET @strpos = @strpos - 1
BREAK
END
ELSE
BEGIN
SET @charpos = @charpos - 1
END
END
IF @charpos = 0 BREAK
END
RETURN SUBSTRING(@input, 1, @strpos)
END
Usage
SELECT dbo.RTRIMCHARS('bla%123', '123%') -- 'bla'
SELECT dbo.RTRIMCHARS('bla%123', '123') -- 'bla%'
SELECT dbo.RTRIMCHARS('bla%123', 'xyz') -- 'bla%123'
SELECT dbo.RTRIMCHARS('bla%123', ']') -- 'bla%123'
SELECT dbo.RTRIMCHARS('bla%123', '') -- 'bla%123'
SELECT dbo.RTRIMCHARS('bla%123', NULL) -- NULL
SELECT dbo.RTRIMCHARS(NULL, '123') -- NULL
I found this document on MS Connect:
http://connect.microsoft.com/SQLServer/feedback/details/259534/patindex-missing-escape-clause
The user asks about ESCAPE
clause with PATINDEX
, then another user extends the request for CHARINDEX
as well.
MS answer: Ticket closed as Won't fix :(
I finished writing my own custom function for LTrim
:
CREATE FUNCTION LTrim_Chars (
@BaseString varchar(2000),
@TrimChars varchar(100)
)
RETURNS varchar(2000) AS
BEGIN
DECLARE @TrimCharFound bit
DECLARE @BaseStringPos int
DECLARE @TrimCharsPos int
DECLARE @BaseStringLen int
DECLARE @TrimCharsLen int
IF @BaseString IS NULL OR @TrimChars IS NULL
BEGIN
RETURN NULL
END
SET @BaseStringPos = 1
SET @BaseStringLen = LEN(@BaseString)
SET @TrimCharsLen = LEN(@TrimChars)
WHILE @BaseStringPos <= @BaseStringLen
BEGIN
SET @TrimCharFound = 0
SET @TrimCharsPos = 1
WHILE @TrimCharsPos <= @TrimCharsLen
BEGIN
IF SUBSTRING(@BaseString, @BaseStringPos, 1) = SUBSTRING(@TrimChars, @TrimCharsPos, 1)
BEGIN
SET @TrimCharFound = 1
BREAK
END
SET @TrimCharsPos = @TrimCharsPos + 1
END
IF @TrimCharFound = 0
BEGIN
RETURN SUBSTRING(@BaseString, @BaseStringPos, @BaseStringLen - @BaseStringPos + 1)
END
SET @BaseStringPos = @BaseStringPos + 1
END
RETURN ''
END
And for RTrim
:
CREATE FUNCTION RTrim_Chars (
@BaseString varchar(2000),
@TrimChars varchar(100)
)
RETURNS varchar(2000) AS
BEGIN
RETURN REVERSE(LTrim_Chars(REVERSE(@BaseString), @TrimChars))
END
At least, I learnt some MsSql scripting...
EDIT:
I added NULL
checks for the two arguments, to reflect Oracle and Postgres' behavior.
Unfortunately, Oracle still behaves slightly differently:
in the case you write LTRIM(string, '')
, it returns NULL
, since a 0-length string is like NULL
in Oracle, so it's actually returning the result of LTRIM(string, NULL)
, which is NULL
indeed.
BTW, this is a really strange case.