avatar
d*o
1
I have files
a.out b.out c.out
How can I copy them to
a.bak b.bak c.bak
or
a.out.bak b.out.bak c.out.bak?
Thanks a lot.
avatar
A*u
2
????
cp a.out a.out.bak
or
mv a.out a.out.bak

【在 d**o 的大作中提到】
: I have files
: a.out b.out c.out
: How can I copy them to
: a.bak b.bak c.bak
: or
: a.out.bak b.out.bak c.out.bak?
: Thanks a lot.

avatar
p*a
3
ls -1 *.out | xargs -i echo {} {} | sed 's/out$/bak/' | xargs -n2 mv

【在 d**o 的大作中提到】
: I have files
: a.out b.out c.out
: How can I copy them to
: a.bak b.bak c.bak
: or
: a.out.bak b.out.bak c.out.bak?
: Thanks a lot.

avatar
d*o
4

Thanks. But is there any much simpler version like using "find ..."?

【在 p*a 的大作中提到】
: ls -1 *.out | xargs -i echo {} {} | sed 's/out$/bak/' | xargs -n2 mv
avatar
j*y
5
for n in `ls *.out`; do cp $n $n.bak; done
you dont have to invoke so many applications :P

【在 d**o 的大作中提到】
:
: Thanks. But is there any much simpler version like using "find ..."?

avatar
m*h
6
//admire

【在 p*a 的大作中提到】
: ls -1 *.out | xargs -i echo {} {} | sed 's/out$/bak/' | xargs -n2 mv
相关阅读
logo
联系我们隐私协议©2024 redian.news
Redian新闻
Redian.news刊载任何文章,不代表同意其说法或描述,仅为提供更多信息,也不构成任何建议。文章信息的合法性及真实性由其作者负责,与Redian.news及其运营公司无关。欢迎投稿,如发现稿件侵权,或作者不愿在本网发表文章,请版权拥有者通知本网处理。