谁能推荐一个安全的解决方案来递归地用下划线替换从给定根目录开始的文件和目录名中的空格?例如:

$ tree
.
|-- a dir
|   `-- file with spaces.txt
`-- b dir
    |-- another file with spaces.txt
    `-- yet another file with spaces.pdf

就变成:

$ tree
.
|-- a_dir
|   `-- file_with_spaces.txt
`-- b_dir
    |-- another_file_with_spaces.txt
    `-- yet_another_file_with_spaces.pdf

当前回答

奈迪姆答案的递归版本。

find . -name "* *" | awk '{ print length, $0 }' | sort -nr -s | cut -d" " -f2- | while read f; do base=$(basename "$f"); newbase="${base// /_}"; mv "$(dirname "$f")/$(basename "$f")" "$(dirname "$f")/$newbase"; done

其他回答

使用rename(又名prename),这是一个Perl脚本,可能已经在您的系统上了。分两步:

find . -name "* *" -type d | rename 's/ /_/g'    # do the directories first
find . -name "* *" -type f | rename 's/ /_/g'

基于Jürgen的回答,能够使用“Revision 1.5 1998/12/18 16:16:31 rmb1”版本的/usr/bin/rename (Perl脚本)在单一边界内处理多层文件和目录:

find /tmp/ -depth -name "* *" -execdir rename 's/ /_/g' "{}" \;

bash 4.0

#!/bin/bash
shopt -s globstar
for file in **/*\ *
do 
    mv "$file" "${file// /_}"       
done

In一样

就像选择的答案一样。

brew install rename

# 
cd <your dir>
find . -name "* *" -type d | rename 's/ /_/g'    # do the directories first
find . -name "* *" -type f | rename 's/ /_/g'

我对这个问题的解决方案是一个bash脚本:

#!/bin/bash
directory=$1
cd "$directory"
while [ "$(find ./ -regex '.* .*' | wc -l)" -gt 0 ];
do filename="$(find ./ -regex '.* .*' | head -n 1)"
mv "$filename" "$(echo "$filename" | sed 's|'" "'|_|g')"
done

只需在执行脚本后将您想要应用脚本的目录名称作为参数。

你可以用这个:

find . -depth -name '* *' | while read fname 

do
        new_fname=`echo $fname | tr " " "_"`

        if [ -e $new_fname ]
        then
                echo "File $new_fname already exists. Not replacing $fname"
        else
                echo "Creating new file $new_fname to replace $fname"
                mv "$fname" $new_fname
        fi
done