Re: Remove Carriage Return
From: Mike Hodgson (mike.hodgson_at_mallesons.nospam.com)
Date: 02/22/05
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Date: Tue, 22 Feb 2005 17:59:41 +1100
Have you tried
replace(t1.address, char(13)+char(10), char(32))
as Mark suggested? It works for me (see example below).
declare @tmp varchar(100)
set @tmp = 'line1
line2
line3
line4
line5'
select @tmp
select replace(@tmp, char(13)+char(10), char(32))
-- *mike hodgson* |/ database administrator/ | mallesons stephen jaques *T* +61 (2) 9296 3668 |* F* +61 (2) 9296 3885 |* M* +61 (408) 675 907 *E* mailto:mike.hodgson@mallesons.nospam.com |* W* http://www.mallesons.com Peter wrote: >Does it mean that I have to run like this > >replace(replace(t1.address, char(13), char(32)), char(10), >char(32)) ? > >Thanks > > > >>-----Original Message----- >>try removing the line feed character as well - char(10) >> >> >> >>regards, >>Mark Baekdal >>http://www.dbghost.com >>http://www.innovartis.co.uk >>+44 (0)208 241 1762 >>Database change management for SQL Server >> >> >> >>"Peter" wrote: >> >> >> >>>I attempt to replace the Carriage Return with Space. >>> >>> >In > > >>>this way, in the SQL Statement, I add this script: >>> >>>replace(t1.address, char(13), char(32)) >>> >>>However, it seems that it only remove the first >>> >>> >carriage > > >>>return but not all. Is there any suggestion ? >>> >>>Thanks >>> >>> >>> >>. >> >> >>
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