update
resource
SET
html_code = REPLACE(REPLACE(html_code, CHAR(10), ''), CHAR(13), ''),
`name` = REPLACE(REPLACE(name, CHAR(10), ''), CHAR(13), '')
where id = 6256;
char(9) 水平制表符 char(10) 换行 char(13) 回车
在不同环境通用的查询含有回车换行sql
select
code ,name
from
resource
where
html_code like ('%'+CHAR(10) + '%' )
or html_code like ('%'+CHAR(13)+CHAR(10) + '%')
or html_code like ('%'+CHAR(13)+ '%');
