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python - 如何在python中创建一棵树？

问题描述

我有一个包含 100 个 html 文件的文件夹，我正在尝试创建一个名称为 page1 - page100 的树。每个页面都有一个超链接，可以打开另一个页面。我试图让它到达程序读取根节点（page1.html）的位置，读取其超链接并根据这些链接创建子节点，然后对其余节点重复此操作，直到树完成。使用超链接解决此问题的最佳方法是什么？到目前为止，这是我的代码。

    import os
from math import* 
from os.path import isfile, join

entries = os.listdir("C:/Users/deonh/Downloads/intranets/intranet1") #This reads the directory

onlyfiles = [f for f in entries if isfile(join("C:/Users/deonh/Downloads/intranets/intranet1", f))] #This took all the webpages in the directory and put them into a list.

print(onlyfiles)

web = open("C:/Users/deonh/Downloads/intranets/intranet1" + "/" + onlyfiles[0]) # This will tell us if the webpage is readable or not

print(web.readable()) # This tells if the file is readable 

print(web.readlines()) #This reads the content of the file

web.close()

标签： python

解决方案

您可以使用 os.walk 遍历所有子目录：

import os
path = 'C:/Users/deonh/Downloads/intranets/intranet1'
entries = []
onlyfiles = []
for directory, _, files in os.walk(path):  # second value would be subdirectries but we don't need them because we iterate over all directories
    entries.append(directory)
    for file in files:
        onlyfiles.append('{}/{}'.format(directory, file))  # 

if os.access(onlyfiles[0], os.R_OK):  # This will tell us if the webpage is readable or not
    with open(onlyfiles[0]) as file:  # file is closing automatically
        print(file.readlines())

print(onlyfiles[0], os.R_OK)  # This tells if the file is readable
print(onlyfiles)  # This reads the content of the file

请询问是否有不清楚的地方。

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