最近接手了一个项目, 图片素材全部是中文命名, 为了避开起名的困扰, 我写了一个脚本,问题迎刃而解, 这里分享给大家
import os
import re
import copy
import shutil
from googletrans import Translator
# 获取当前目录下所有的css文件路径
def getAllMd (file_dir):
all_whole_path_files = []
for root, dirs, files in os.walk(file_dir):
for file in files:
try:
if (file[-4:] == ".png") or (file[-4:] == ".jpg"):
file_info = [root+'/', file]
all_whole_path_files.append(file_info)
except Exception as e:
print(e)
return all_whole_path_files
# 将中文转换为英文
def getRepName(path_file):
# 暂时保留后缀
extension_name = ''
extension_name = path_file[1].split(".")[-1]
# 无后缀的文件名
path_file[1] = path_file[1][0:-len(extension_name)-1]
# 实例化翻译
translator = Translator()
tmp_en_name = translator.translate(path_file[1], dest='en').text
# 将数字和字母保存到最终的字符串中, 遇到空格则替换为中划线保存
en_name = ''
for en_name_str in tmp_en_name:
# 将大写字母转换为小写字母
en_name_str = en_name_str.lower()
# 保留小写字母
if re.match('[a-z]', en_name_str):
en_name += en_name_str
# 将空格转换为"-"
elif en_name_str == " ":
en_name += "-"
else:
pass
# 补充后缀名
path_file[1] = en_name + '.' + extension_name
return path_file
# 根据相对路径及新旧英文名 创建新文件
def createNewFile(whole_path_file, new_whole_path_file):
# 在图片的同级目录, 创建一个enName的文件夹
if os.path.exists(whole_path_file[0] + "./enName/"):
pass
else:
os.makedirs(whole_path_file[0]+ "./enName/")
# 拷贝创建新的文件
shutil.copyfile(whole_path_file[0]+whole_path_file[1], new_whole_path_file[0]+"./enName/"+new_whole_path_file[1])
def main():
all_whole_path_files = getAllMd('./')
for whole_path_file in all_whole_path_files:
# 获取英文名
new_whole_path_file = getRepName(copy.deepcopy(whole_path_file))
print("旧的路径和英文名:", whole_path_file, "新的路径和英文名:", new_whole_path_file)
# 根据相对路径及新旧英文名创建新文件
createNewFile(whole_path_file, new_whole_path_file)
if __name__ == '__main__':
main()
这个脚本只是解决了我的问题, 如果有新的需求, 欢迎到这个脚本对应的Github提交需求, 也欢迎点个星, https://github.com/zhaoolee/zhToEnName