如何使用“变量”批量重命名文件？

Question 1

如果您足够热衷于修改正则表达式（在提供的示例中），那么我建议您更进一步，编写一个小脚本 - 比如，用 Python。这样，您将能够对文件名执行任何转换。

我估计 Python 脚本不会超过 15-20 行，所以这绝对不是一项艰巨的任务。仅使用正则表达式（正如您所尝试的那样）会受到更多限制。

以下是我对此类脚本的看法：

#!/usr/bin/python
import os,re

files = os.listdir('.')

SEASONS = (
 (1, 1, 3), # the format is - season number, first episode, last episode
 (2, 50,52),
 (3, 53,55),
 (4, 56,99),
)

for f in files:
    # skip all files which are not .mp4
    if not f.endswith(".mp4"):
        continue

    # find the first number in the filename
    matches = re.findall("\d+", f)
    if not len(matches):
       print "skipping", f
    num = int(matches[0])

    for season in SEASONS:
        if num <= season[2]:
            season_num = season[0]
            ep_num = num - season[1] + 1
            new_file_name = "BleachS%02dE%02d.mp4" % (season_num, ep_num)
            # This is for testing
            print "%s ==> %s" % (f, new_file_name)
            # Uncomment the following when you're satisfied with the test runs
            # os.rename(f, new_file_name)
            break

print "Done"

看起来我低估了脚本的大小（目前是 36 行），不过我确信如果你带着这段代码去 stackoverflow，你会得到很多更优雅的建议

我说了它可以在 15 行内完成……下面是 20 行，其中 5 行是配置 :P

#!/usr/bin/python
import os, re, glob

SEASONS = (
 {'num':1, 'first':1, 'last':3}, # the format is - season number, first episode, last episode
 {'num':2, 'first':50, 'last':52},
 {'num':3, 'first':53, 'last':55},
 {'num':4, 'first':56, 'last':99},
)

files = glob.glob('bleach*.mp4')
for f in files:
    num = int(re.findall("\d+", f)[0])  # find the first number in the filename
    for season in SEASONS:
        if num > season['last']: continue
        new_file_name = "BleachS%02dE%02d.mp4" % (season['num'], num - season['first'] + 1)
        print "%s ==> %s" % (f, new_file_name) # This is for testing
        # os.rename(f, new_file_name) # Uncomment this when you're satisfied with the test runs
        break

Answer

如果您足够热衷于修改正则表达式（在提供的示例中），那么我建议您更进一步，编写一个小脚本 - 比如，用 Python。这样，您将能够对文件名执行任何转换。

我估计 Python 脚本不会超过 15-20 行，所以这绝对不是一项艰巨的任务。仅使用正则表达式（正如您所尝试的那样）会受到更多限制。

以下是我对此类脚本的看法：

#!/usr/bin/python
import os,re

files = os.listdir('.')

SEASONS = (
 (1, 1, 3), # the format is - season number, first episode, last episode
 (2, 50,52),
 (3, 53,55),
 (4, 56,99),
)

for f in files:
    # skip all files which are not .mp4
    if not f.endswith(".mp4"):
        continue

    # find the first number in the filename
    matches = re.findall("\d+", f)
    if not len(matches):
       print "skipping", f
    num = int(matches[0])

    for season in SEASONS:
        if num <= season[2]:
            season_num = season[0]
            ep_num = num - season[1] + 1
            new_file_name = "BleachS%02dE%02d.mp4" % (season_num, ep_num)
            # This is for testing
            print "%s ==> %s" % (f, new_file_name)
            # Uncomment the following when you're satisfied with the test runs
            # os.rename(f, new_file_name)
            break

print "Done"

看起来我低估了脚本的大小（目前是 36 行），不过我确信如果你带着这段代码去 stackoverflow，你会得到很多更优雅的建议

我说了它可以在 15 行内完成……下面是 20 行，其中 5 行是配置 :P

#!/usr/bin/python
import os, re, glob

SEASONS = (
 {'num':1, 'first':1, 'last':3}, # the format is - season number, first episode, last episode
 {'num':2, 'first':50, 'last':52},
 {'num':3, 'first':53, 'last':55},
 {'num':4, 'first':56, 'last':99},
)

files = glob.glob('bleach*.mp4')
for f in files:
    num = int(re.findall("\d+", f)[0])  # find the first number in the filename
    for season in SEASONS:
        if num > season['last']: continue
        new_file_name = "BleachS%02dE%02d.mp4" % (season['num'], num - season['first'] + 1)
        print "%s ==> %s" % (f, new_file_name) # This is for testing
        # os.rename(f, new_file_name) # Uncomment this when you're satisfied with the test runs
        break

Question 2

我给你写了一个python脚本

import sys
import os
import glob
import re

def rename():
    episode_counts = (0, 20, 41, 63)
    episode_pairs = []
    for index, num in enumerate(episode_counts):
        if index < len(episode_counts) - 1:
            episode_pairs.append((num, episode_counts[index+1]))

    episodes = glob.glob(os.path.join(sys.argv[1], '*'))

    for episode in episodes:
        match = re.search('.*(\d{2}).*$', os.path.basename(episode))
        episode_num = match.group(1)

        for season, pair in enumerate(episode_pairs):
            if int(episode_num) in range(pair[0]+1, pair[1]+1):
                dirname = os.path.dirname(episode)
                path = os.path.join(dirname, 'Bleach S{0}E{1}'.format(season+1, episode_num))
                os.rename(episode, path)

if __name__ == "__main__":
    rename()

我对 Python 还很陌生（这也是我写这篇文章的原因之一，为了练习），所以它可能不是世界上最好的脚本。但我在一些测试文件上试了一下，似乎有效。

只需将靠近顶部的行编辑episode_counts = ...为该季的最后一集数。我输入了前三集这一页。

将代码保存为类似的内容episode_renamer.py并将其与一起使用python episode_renamer.py /path/to/episodes/。

Answer

我给你写了一个python脚本

import sys
import os
import glob
import re

def rename():
    episode_counts = (0, 20, 41, 63)
    episode_pairs = []
    for index, num in enumerate(episode_counts):
        if index < len(episode_counts) - 1:
            episode_pairs.append((num, episode_counts[index+1]))

    episodes = glob.glob(os.path.join(sys.argv[1], '*'))

    for episode in episodes:
        match = re.search('.*(\d{2}).*$', os.path.basename(episode))
        episode_num = match.group(1)

        for season, pair in enumerate(episode_pairs):
            if int(episode_num) in range(pair[0]+1, pair[1]+1):
                dirname = os.path.dirname(episode)
                path = os.path.join(dirname, 'Bleach S{0}E{1}'.format(season+1, episode_num))
                os.rename(episode, path)

if __name__ == "__main__":
    rename()

我对 Python 还很陌生（这也是我写这篇文章的原因之一，为了练习），所以它可能不是世界上最好的脚本。但我在一些测试文件上试了一下，似乎有效。

只需将靠近顶部的行编辑episode_counts = ...为该季的最后一集数。我输入了前三集这一页。

将代码保存为类似的内容episode_renamer.py并将其与一起使用python episode_renamer.py /path/to/episodes/。

Question 3

存储库中有一个名为 Thunar 的文件管理器，它具有批量重命名选项。

请参阅此网站以了解更多信息： http://thunar.xfce.org/pwiki/documentation/bulk_renamer

您可以在软件中心或命令行中安装 thunar：

sudo apt-get install thunar

在 bash 中还有其他方法可以做到这一点，或者通过编写一些代码，但图形工具可能会更快地让您得到您想要的结果。

Answer