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

我需要在Windows 7上的Cygwin设置上这样做,并发现我在上面尝试的建议中出现了语法错误(尽管我可能错过了一个工作选项)。然而,这个解决方案直接来自Ubuntu论坛:-)

ls | while read upName; do loName=`echo "${upName}" | tr '[:upper:]' '[:lower:]'`; mv "$upName" "$loName"; done

(注意:我之前用下划线替换了空格:

for f in *\ *; do mv "$f" "${f// /_}"; done

)

其他回答

for f in `find`; do mv -v "$f" "`echo $f | tr '[A-Z]' '[a-z]'`"; done

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])

下面是我的次优解决方案,使用Bash shell脚本:

#!/bin/bash
# First, rename all folders
for f in `find . -depth ! -name CVS -type d`; do
   g=`dirname "$f"`/`basename "$f" | tr '[A-Z]' '[a-z]'`
   if [ "xxx$f" != "xxx$g" ]; then
      echo "Renaming folder $f"
      mv -f "$f" "$g"
   fi
done

# Now, rename all files
for f in `find . ! -type d`; do
   g=`dirname "$f"`/`basename "$f" | tr '[A-Z]' '[a-z]'`
   if [ "xxx$f" != "xxx$g" ]; then
      echo "Renaming file $f"
      mv -f "$f" "$g"
   fi
done

文件夹都正确地重命名,mv在权限不匹配时不会询问问题,CVS文件夹也不会重命名(不幸的是,该文件夹内的CVS控制文件仍然会重命名)。

由于“find -depth”和“find | sort -r”都以可用于重命名的顺序返回文件夹列表,我更喜欢使用“-depth”来搜索文件夹。

这里的解决方案都不适合我,因为我所在的系统无法访问perl重命名脚本,而且有些文件包含空格。然而,我发现了一个有效的变体:

find . -depth -exec sh -c '
    t=${0%/*}/$(printf %s "${0##*/}" | tr "[:upper:]" "[:lower:]");
    [ "$t" = "$0" ] || mv -i "$0" "$t"
' {} \;

归功于“Gilles 'SO-停止作恶”,在Unix和Linux StackExchange上看到类似问题“将整个目录树更改为小写名称”的答案。

MacOS,

安装重命名包,

brew install rename

Use,

find . -iname "*.py" -type f | xargs -I% rename -c -f  "%"                       

该命令查找所有扩展名为*.py的文件,并将文件名转换为小写。

`f` - forces a rename

例如,

$ find . -iname "*.py" -type f
./sample/Sample_File.py
./sample_file.py
$ find . -iname "*.py" -type f | xargs -I% rename -c -f  "%"
$ find . -iname "*.py" -type f
./sample/sample_file.py
./sample_file.py