1、生成项目目录树
在当前文件所在文件夹下运行。
2、代码
dir_tree.py
# -*- coding: utf-8 -*-
import sys
from pathlib import Path
class DirectionTree(object):
"""生成目录树
@ pathname: 目标目录
@ filename: 要保存成文件的名称
"""
def __init__(self, pathname='.', filename='tree.txt'):
super(DirectionTree, self).__init__()
self.pathname = Path(pathname)
self.filename = filename
self.tree = ''
def set_path(self, pathname):
self.pathname = Path(pathname)
def set_filename(self, filename):
self.filename = filename
def generate_tree(self, n=0):
if self.pathname.is_file():
self.tree += ' |' * n + '-' * 4 + self.pathname.name + '\n'
elif self.pathname.is_dir():
self.tree += ' |' * n + '-' * 4 + \
str(self.pathname.relative_to(self.pathname.parent)) + '\\' + '\n'
for cp in self.pathname.iterdir():
self.pathname = Path(cp)
self.generate_tree(n + 1)
def save_file(self):
with open('./'+self.filename, 'w', encoding='utf-8') as f:
f.write(self.tree)
print("已保存文件!")
if __name__ == '__main__':
# 运行
# python dir_tree.py D:\Work\SlabSerialRecognition\MyCRNN_Slab dir_tree.txt
dirtree = DirectionTree()
# 命令参数个数为1,生成当前目录的目录树
if len(sys.argv) == 1:
dirtree.set_path(Path.cwd())
dirtree.generate_tree()
dirtree.save_file()
print(dirtree.tree)
# 命令参数个数为2并且目录存在存在
elif len(sys.argv) == 2 and Path(sys.argv[1]).exists():
dirtree.set_path(sys.argv[1])
dirtree.generate_tree()
print(dirtree.tree)
# 命令参数个数为3并且目录存在存在
elif len(sys.argv) == 3 and Path(sys.argv[1]).exists():
dirtree.set_path(sys.argv[1])
dirtree.generate_tree()
dirtree.set_filename(sys.argv[2])
dirtree.save_file()
else: # 参数个数太多,无法解析
print('命令行参数太多,请检查!')
标签:Python,self,tree,生成,sys,pathname,filename,目录,dirtree
From: https://www.cnblogs.com/jackchen28/p/18641495