如果您在UNIX上有
rename
命令,您应该能够使用类似以下的命令:
mkdir images_renamed
cd images_renamed
cp ../103*.jpg .
rename 103 201003 *.jpg
rename 's/FROM/TO/' FILE
命令可以将
FILE
中所有指定的文件重命名,将第一次出现的
FROM
替换为
TO
。
如果没有这个命令,你可以使用类似以下的命令:
mkdir images_renamed
for fspec in 103*.jpg ; do
cp ${fspec} images_renamed/201003${fspec:3}
done
为了递归地完成这个任务,我会将其放入一个带有
find
命令的脚本中:
#!/usr/bin/bash
rm -rf images_renamed
ls -lR images
echo
cd images
find . -name '*.jpg' | while read -r; do
mkdir -p "../images_renamed/$(dirname "$REPLY")"
echo 'Copying from' [$REPLY]
echo ' to' [../images_renamed/$REPLY] and renaming.
echo
cp "$REPLY" "../images_renamed/$REPLY"
cd "$(dirname "../images_renamed/$REPLY")"
rename 103 201003 "$(basename "$REPLY")"
cd - >/dev/null
done
cd ..
ls -lR images_renamed
只有中间部分是必需的,其余部分是用于测试的。下面的输出显示了它的工作原理,将每个文件复制到新的目录结构并重命名相关文件。
images:
total 0
drwxr-xr-x+ 1 pax None 0 2010-08-12 20:55 dir1
drwxr-xr-x+ 1 pax None 0 2010-08-12 20:55 dir2
drwxr-xr-x+ 1 pax None 0 2010-08-12 20:56 dir3
images/dir1:
total 0
-rw-r--r-- 1 pax None 0 2010-08-12 20:55 102xxx.jpg
-rw-r--r-- 1 pax None 0 2010-08-12 20:55 103xxx.jpg
images/dir2:
total 0
-rw-r--r-- 1 pax None 0 2010-08-12 20:55 103yyy.jpg
images/dir3:
total 0
drwxr-xr-x+ 1 pax None 0 2010-08-12 20:55 dir 4
images/dir3/dir 4:
total 0
-rw-r--r-- 1 pax None 0 2010-08-12 20:55 103zzz.jpg
Copying from [./dir1/102xxx.jpg]
to [../images_renamed/./dir1/102xxx.jpg] and renaming.
Copying from [./dir1/103xxx.jpg]
to [../images_renamed/./dir1/103xxx.jpg] and renaming.
Copying from [./dir2/103yyy.jpg]
to [../images_renamed/./dir2/103yyy.jpg] and renaming.
Copying from [./dir3/dir 4/103zzz.jpg]
to [../images_renamed/./dir3/dir 4/103zzz.jpg] and renaming.
images_renamed:
total 0
drwxr-xr-x+ 1 pax None 0 2010-08-12 21:19 dir1
drwxr-xr-x+ 1 pax None 0 2010-08-12 21:19 dir2
drwxr-xr-x+ 1 pax None 0 2010-08-12 21:19 dir3
images_renamed/dir1:
total 0
-rw-r--r-- 1 pax None 0 2010-08-12 21:19 102xxx.jpg
-rw-r--r-- 1 pax None 0 2010-08-12 21:19 201003xxx.jpg
images_renamed/dir2:
total 0
-rw-r--r-- 1 pax None 0 2010-08-12 21:19 201003yyy.jpg
images_renamed/dir3:
total 0
drwxr-xr-x+ 1 pax None 0 2010-08-12 21:19 dir 4
images_renamed/dir3/dir 4:
total 0
-rw-r--r-- 1 pax None 0 2010-08-12 21:19 201003zzz.jpg
为了扁平化文件层次结构,您可以使用类似以下的内容:
#!/usr/bin/bash
rm -rf images_renamed
ls -lR images
echo
cd images
mkdir -p ../images_renamed
find . -name '*.jpg' | while read -r; do
newfile="$(basename "$REPLY")"
echo 'Copying from' [$REPLY]
echo ' to' [../images_renamed/$newfile] and renaming.
echo
cp "$REPLY" "../images_renamed/$newfile"
cd ../images_renamed
rename 103 201003 "$newfile"
cd - >/dev/null
done
这将输出:
cd ..
ls -lR images_renamed
images:
total 0
drwxr-xr-x+ 1 allan None 0 2010-08-12 20:55 dir1
drwxr-xr-x+ 1 allan None 0 2010-08-12 20:55 dir2
drwxr-xr-x+ 1 allan None 0 2010-08-12 20:56 dir3
images/dir1:
total 0
-rw-r--r-- 1 allan None 0 2010-08-12 20:55 102xxx.jpg
-rw-r--r-- 1 allan None 0 2010-08-12 20:55 103xxx.jpg
images/dir2:
total 0
-rw-r--r-- 1 allan None 0 2010-08-12 20:55 103yyy.jpg
images/dir3:
total 0
drwxr-xr-x+ 1 allan None 0 2010-08-12 20:55 dir 4
images/dir3/dir 4:
total 0
-rw-r--r-- 1 allan None 0 2010-08-12 20:55 103zzz.jpg
Copying from [./dir1/102xxx.jpg]
to [../images_renamed/102xxx.jpg] and renaming.
Copying from [./dir1/103xxx.jpg]
to [../images_renamed/103xxx.jpg] and renaming.
Copying from [./dir2/103yyy.jpg]
to [../images_renamed/103yyy.jpg] and renaming.
Copying from [./dir3/dir 4/103zzz.jpg]
to [../images_renamed/103zzz.jpg] and renaming.
images_renamed:
total 0
-rw-r--r-- 1 allan None 0 2010-08-12 22:41 102xxx.jpg
-rw-r--r-- 1 allan None 0 2010-08-12 22:41 201003xxx.jpg
-rw-r--r-- 1 allan None 0 2010-08-12 22:41 201003yyy.jpg
-rw-r--r-- 1 allan None 0 2010-08-12 22:41 201003zzz.jpg
但是需要记住的是,文件名冲突(不同目录下相同的文件名)会互相覆盖。
rename
脚本可以轻松完成此任务。实际上,编写Perl的动机之一是awk对许多文件的有限支持。rename
应该已经是基于debian的发行版或http://search.cpan.org/~rmbarker/File-Rename-0.05/的一部分。 - msw