bash 在 Linux 中重命名多个文件 shell
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/11108735/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
rename multiple files shell in Linux
提问by Linux Rules
I have a number of files such as file_022.bmp, file_023.bmp...file_0680.bmp. I need to rename these to something a little bit more convenient such as file_1.bmp, file_2.bmp...file_658.bmp.
我有许多文件，例如 file_022.bmp、file_023.bmp...file_0680.bmp。我需要将它们重命名为更方便的名称，例如 file_1.bmp、file_2.bmp...file_658.bmp。
Is there a bash script that I could write to do this for me? Thanks for the help and advice.
是否有我可以编写的 bash 脚本来为我执行此操作？感谢您的帮助和建议。
Luke H
卢克·H
回答by Harald Brinkhof
if you're on a debian based linux system then you can use the rename script which accepts regular expressionsto rename files. Some more infobecause I find it hard to find the man page.
如果您使用的是基于 debian 的 linux 系统，那么您可以使用接受正则表达式的重命名脚本来重命名文件。更多信息，因为我发现很难找到手册页。
e.g.
例如
harald@Midians_Gate:~$ ls p*.php
parse.php pd.php pgrep.php preg_based.php proc.php
suppose I want to change the extension to .perl and prepend the name with file_ then I use command:
假设我想将扩展名更改为 .perl 并在名称前加上 file_ 然后我使用命令：
rename -n 's/([a-z]*)\.php/file_.perl/' p*.php
would give
会给
parse.php renamed as file_parse.perl
pd.php renamed as file_pd.perl
pgrep.php renamed as file_pgrep.perl
preg_based.php renamed as preg_file_based.perl
proc.php renamed as file_proc.perl
I select and capture the base filename ([a-z]*)and then use it in the substitution $1and append .perland prepend $1 with the regular string file_
我选择并捕获基本文件名([a-z]*)，然后在替换中使用它，$1并在.perl$1 前面加上常规字符串file_
the -n option makes it test run without changing anything
-n 选项使其在不更改任何内容的情况下测试运行
As you can see from this example your selecting regexp needs to be correctly thought out or you get cases like the above preg_based.php where you wanted file_preg_based.perl :)
to compensate for that I would've needed to use ([a-z_]*)here
正如你从这个例子中看到的，你需要正确地考虑选择正则表达式，或者你会得到像上面的 preg_based.php 这样的情况，你想要 file_preg_based.perl :) 来补偿我需要在([a-z_]*)这里使用
It's one of the many reasons why I keep hanging on to debian, I'd love to find the equivalent for other non-debian systems though :-/
这是我一直坚持使用 debian 的众多原因之一，不过，我很想找到其他非 debian 系统的等价物：-/
回答by stelios
if you have files a.bmp,b.bmp,c.bmp and you want to end up with file_1.bmp, file_2.bmp, file_3.bmp
如果你有文件 a.bmp、b.bmp、c.bmp 并且你想以 file_1.bmp、file_2.bmp、file_3.bmp 结束
using bash:
使用 bash：
mkdir result
index=1
for i in *.bmp
do
mv "$i" "result/file_"$((index++)).bmp
done
notes: using a subdirectory is advised to avoid accidentally overwriting a file that looks like file_xx.bmp
注意：建议使用子目录以避免意外覆盖看起来像 file_xx.bmp 的文件
if you have too many files to fit in the command line after expansion you could use something like:
如果扩展后有太多文件无法放入命令行，则可以使用以下内容：
mkdir result
index=1
find . -name "*.bmp" | while read i
do
echo mv "$i" "result/file_"$((index++)).bmp
done
after inspecting the output remove the 'echo'
检查输出后删除“回声”
