select * from dbo.luDefect
update luDefect
set Description= REPLACE(Description, CHAR(13), '')
where REPLACE(Description, CHAR(13), '<br>') like '%<br>%'
select * from dbo.luDefect
update luDefect
set Description= REPLACE(Description, CHAR(13), '')
where REPLACE(Description, CHAR(13), '<br>') like '%<br>%'