小弟在此等候回答。
update data set description
=
select left(description,charindex( ' <Br> ',description) -1) as description from data where num in(
select num from data
where description like '% <Br> % ')
where where description like '% <Br> % '
我的意思是，把description修改成查询出的 ,但是我这样写不可以，为什么？
[解决办法]
Try:
update data set description
=left(description,charindex( ' <Br> ',description) -1)
where description like '% <Br> % '
