bash 在一行 shell 中重命名多个文件
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Renaming multiple files in one line of shell
提问by Susam Pal
Problem
问题
In a directory there are files of the format: *-foo-bar.txt.
在一个目录中有格式为：*-foo-bar.txt 的文件。
Example directory:
示例目录：
$ ls *-*
asdf-foo-bar.txt ghjk-foo-bar.txt l-foo-bar.txt tyui-foo-bar.txt
bnm-foo-bar.txt iop-foo-bar.txt qwer-foo-bar.txt zxcv-foo-bar.txt
Desired directory:
所需目录：
$ ls *.txt
asdf.txt bnm.txt ghjk.txt iop.txt l.txt qwer.txt tyui.txt zxcv.txt
Solution 1
方案一
The first solution that came to my mind looks somewhat like this ugly hack:
我想到的第一个解决方案看起来有点像这个丑陋的黑客：
ls *-* | cut -d- -f1 | sed 's/.*/mv "for i in *-*
do
mv "$i" "`echo $i | cut -f1 -d-`.txt"
done
-foo-bar.txt" "rename 's/-foo-bar//' *-foo-bar.txt
.txt"/' > rename.sh && sh rename.sh
The above solution creates a script, on the fly, to rename the files. This solution also tries to parse the output of lswhich is not a good thing to do as per http://mywiki.wooledge.org/ParsingLs.
上述解决方案会动态创建一个脚本来重命名文件。此解决方案还尝试解析输出，ls根据http://mywiki.wooledge.org/ParsingLs这样做不是一件好事。
Solution 2
解决方案2
This problem can be solved more elegantly with a shell script like this:
这个问题可以用这样的 shell 脚本更优雅地解决：
find . -maxdepth 1 -mindepth 1 -type f -name '*-foo-bar.txt' | sed 's/-foo-bar.txt//' | xargs -I{} mv {}-foo-bar.txt {}.txt
The above solution uses a loop to rename the files.
上述解决方案使用循环重命名文件。
Question
题
Is there a way to solve this problem in a single line such that we do not have to explicitly script a loop, or generate a script, or invoke a new or the current shell (i.e. avoid sh, bash, ., etc. commands)?
有没有办法在一行中解决这个问题，这样我们就不必显式编写循环脚本，或生成脚本，或调用新的或当前的 shell（即避免使用 sh、bash、. 等命令） ?
