Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
407 views
in Technique[技术] by (71.8m points)

shell - Process all arguments except the first one (in a bash script)

I have a simple script where the first argument is reserved for the filename, and all other optional arguments should be passed to other parts of the script.

Using Google I found this wiki, but it provided a literal example:

echo "${@: -1}"

I can't get anything else to work, like:

echo "${@:2}"

or

echo "${@:2,1}"

I get "Bad substitution" from the terminal.

What is the problem, and how can I process all but the first argument passed to a bash script?

Question&Answers:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Use this:

echo "${@:2}"

The following syntax:

echo "${*:2}"

would work as well, but is not recommended, because as @Gordon already explained, that using *, it runs all of the arguments together as a single argument with spaces, while @ preserves the breaks between them (even if some of the arguments themselves contain spaces). It doesn't make the difference with echo, but it matters for many other commands.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...