我必须递归地重命名一个完整的文件夹树,这样就不会出现大写字母(这是c++源代码,但这无关紧要)。

忽略CVS和Subversion版本控制文件/文件夹的加分项。首选的方法是shell脚本,因为shell应该在任何Linux机器上可用。

关于文件重命名的细节有一些有效的争论。

I think files with the same lowercase names should be overwritten; it's the user's problem. When checked out on a case-ignoring file system, it would overwrite the first one with the latter, too. I would consider A-Z characters and transform them to a-z, everything else is just calling for problems (at least with source code). The script would be needed to run a build on a Linux system, so I think changes to CVS or Subversion version control files should be omitted. After all, it's just a scratch checkout. Maybe an "export" is more appropriate.


当前回答

我相信这些一行代码可以简化为:

对于f在**/*;做mv "$f" "${f:l}";完成

其他回答

最初的问题要求忽略SVN和CVS目录,这可以通过在find命令中添加-prune来实现。例如忽略CVS:

find . -name CVS -prune -o -exec mv '{}' `echo {} | tr '[A-Z]' '[a-z]'` \; -print

[编辑]我尝试了一下,在find中嵌入小写翻译并没有起作用,原因我实际上不明白。因此,将其修改为:

$> cat > tolower
#!/bin/bash
mv $1 `echo $1 | tr '[:upper:]' '[:lower:]'`
^D
$> chmod u+x tolower 
$> find . -name CVS -prune -o -exec tolower '{}'  \;

Ian

如果你使用Arch Linux,你可以从AUR安装rename)包,它提供了renamexm命令作为/usr/bin/renamexm可执行文件和一个手册页面。

它是一个非常强大的工具,可以快速重命名文件和目录。

转换为小写字母

rename -l Developers.mp3 # or --lowcase

转换为大写

rename -u developers.mp3 # or --upcase, long option

其他选项

-R --recursive # directory and its children

-t --test # Dry run, output but don't rename

-o --owner # Change file owner as well to user specified

-v --verbose # Output what file is renamed and its new name

-s/str/str2 # Substitute string on pattern

--yes # Confirm all actions

如果需要,您可以从这里获取示例Developers.mp3文件;)

这是一个小的shell脚本,做你所要求的:

root_directory="${1?-please specify parent directory}"
do_it () {
    awk '{ lc= tolower($0); if (lc != $0) print "mv \""  $0 "\" \"" lc "\"" }' | sh
}
# first the folders
find "$root_directory" -depth -type d | do_it
find "$root_directory" ! -type d | do_it

注意第一个find中的-depth动作。

在这种情况下,我会使用Python,以避免乐观地假设没有空格或斜杠的路径。我还发现python2往往被安装在更多的地方,而不是重命名。

#!/usr/bin/env python2
import sys, os

def rename_dir(directory):
  print('DEBUG: rename('+directory+')')

  # Rename current directory if needed
  os.rename(directory, directory.lower())
  directory = directory.lower()

  # Rename children
  for fn in os.listdir(directory):
    path = os.path.join(directory, fn)
    os.rename(path, path.lower())
    path = path.lower()

    # Rename children within, if this child is a directory
    if os.path.isdir(path):
        rename_dir(path)

# Run program, using the first argument passed to this Python script as the name of the folder
rename_dir(sys.argv[1])

使用bash,不重命名:

find . -exec bash -c 'mv $0 ${0,,}' {} \;