显示包含文件的路径树

Question

import os
path = "G:krunkermod"
abcde = open("path.txt", "w")
for dirpath, dirnames, filenames in os.walk(path):
directory_level = dirpath.replace(path, "")
directory_level = directory_level.count(os.sep)
indent = " " * 4
print("{}{}/".format(indent*directory_level, os.path.basename(dirpath)), file=abcde)
for f in filenames:
print("{}{}".format(indent*(directory_level+1), f), file=abcde)
abcde.close()

我想让它打印路径中每个文件夹的文件但它只打印最后一个

Answer 1

缩进不正确。第二个for环也必须在第一个for环内。

正确代码:

import os
path = "/home/user/my_folder/tools"
abcde = open("path.txt", "w")
for dirpath, dirnames, filenames in os.walk(path):
directory_level = dirpath.replace(path, "")
directory_level = directory_level.count(os.sep)
indent = " " * 4
print("{}{}/".format(indent*directory_level, os.path.basename(dirpath)), file=abcde)
for f in filenames:
print("{}{}".format(indent*(directory_level+1), f), file=abcde)
abcde.close()

path.txt内容的一部分:

tools/
.gitignore
README.md
__init__.py
requirements3.txt
test.py
path.txt
.git/
description
hooks/
commit-msg.sample
info/
exclude
refs/
heads/
master

Answer 2

您也可以使用递归函数，递归地获取所有子文件夹的内容，直到存在任何子文件夹:

from os import walk
output = open("path.txt", "w")
def listFiles(path, indent):
for (openedPath, folders, files) in walk(path):
for file in files:
output.write("t" * (indent) + file + "n")
for folder in folders:
output.write("t" * (indent) + folder + "/n")
listFiles(path + "/" + folder, indent + 1)
break
source = "/my/path/to/my/folder"
print(source + "/")
listFiles(source, 1)

有一个小代码项目文件夹的例子。

/my/path/to/my/folder/
input.txt
main.py
output/
error.cpp
trying.cpp
logo.cpp

Answer 3

您可以使用这种代码来简化

import os
folder = r"C:pathtofindfiles"
x = [os.path.join(r,file) for r,d,f in os.walk(folder) for file in f # If you want specific files if file.endswith(".txt")]
y = [os.path.join(r,folder) for r,d,f in os.walk(folder) for folder in d]
print(x) #For files in main directory and subdirectories
print(y) #For files in main directory and subdirectories