Try:
Select * from yourTable where charindex(char(10) + char(13),yourField) > 0
update yourTable set yourField = replace(yourField,char(10) + char(13),'')
Select * from yourTable where charindex(char(10) + char(13),yourField) > 0
update yourTable set yourField = replace(yourField,char(10) + char(13),'')
制表符 CHAR(9)
换行符 CHAR(10)
回车 CHAR(13)
set 要处理的字段= replace(要处理的字段,char(10) + char(13),'')--char(10)+char(13) 即为回车换行符
set 要处理的字段= replace(要处理的字段,char(13) + char(10),'')
--测试过程
declare @tb table(aa varchar(1000))
insert @tb
select 'asdfasdf
asdfjasld
asdfjdf
sdfasdf
asdfasdf'declare @a varchar(1000)
select @a=replace(aa,char(10)+char(13),'') from @tb
print '换行+回车的替换方式'
print @aselect @a=replace(aa,char(13)+char(10),'') from @tb
print '回车+换行的替换方式'
print @a--测试结果(所影响的行数为 1 行)换行+回车的替换方式
asdfasdf
asdfjasld
asdfjdf
sdfasdf
asdfasdf
回车+换行的替换方式
asdfasdf asdfjasldasdfjdf sdfasdfasdfasdf