编写 bash 脚本以按顺序重命名当前目录和子目录中的文件

Question

如何修复此脚本，以便它使用用户输入的 STRING 重命名当前目录中的所有文件以及子目录中的所有文件？

当前文件名可以是任何内容。我希望所有文件的最终结果都被称为STRING_001.jpg STRING_002.jpg STRING_003.jpg等等......

如果文件仅在各自的目录中是连续的，或者文件编号是否转移到下一个目录，这对我来说并不重要。

这是我经过几个小时的研究最终编写的代码：

#!/bin/bash
# Let user input a string to use in filenames
read -p "Enter filename: " STRING
# Variable for sequential filenames
n=0
files=(./*)
# Looks for files in current directory
if [ ${#files[@]} -gt 0 ]; 
then
for f in ./*; do
# Renames files with STRING and sequentially
printf -v num %03d "$((++i))"
mv "$f" "${STRING}_$num.jpg" 
done
else 
for d in *; do
( cd -- "$d" && 
for f in ./*; do 
# Executes rename command in sub-directories
printf -v num %03d "$((++i))"
mv "$f" "${STRING}_$num.jpg" 
done 
)
done;
fi

Answer 1

脚本：

#!/bin/bash
read -p "Enter filename: " STRING
find . -type f | cat -n - | while read num infile ; do
filename=$(basename "${infile}")
filesans="${filename%%.*}"
newname=$(printf '%s_%03d' "${STRING}" "${num}")
newfile="${infile/$filesans/$newname}"
echo "mv "${infile}" "${newfile}""
done

(此处，此脚本名为62566825.sh。不是最伟大的名字。

设置目录：

mkdir dir1 dir2 dir3
touch dir1/file{1,2,3} dir2/file{4,5,6}.jpg dir3/file{7,8,9}.foo.gz
$ find . -type f
./62566825.sh
./dir1/file1
./dir1/file2
./dir1/file3
./dir2/file4.jpg
./dir2/file5.jpg
./dir2/file6.jpg
./dir3/file7.foo.gz
./dir3/file8.foo.gz
./dir3/file9.foo.gz

执行：

$ ./62566825.sh
Enter filename: QUUX
mv "./62566825.sh" "./QUUX_001.sh"
mv "./dir1/file1" "./dir1/QUUX_002"
mv "./dir1/file2" "./dir1/QUUX_003"
mv "./dir1/file3" "./dir1/QUUX_004"
mv "./dir2/file4.jpg" "./dir2/QUUX_005.jpg"
mv "./dir2/file5.jpg" "./dir2/QUUX_006.jpg"
mv "./dir2/file6.jpg" "./dir2/QUUX_007.jpg"
mv "./dir3/file7.foo.gz" "./dir3/QUUX_008.foo.gz"
mv "./dir3/file8.foo.gz" "./dir3/QUUX_009.foo.gz"
mv "./dir3/file9.foo.gz" "./dir3/QUUX_010.foo.gz"

(只需修复echo "mv即可实际执行某些操作。