如何在 python 中創建樹？

Question

我有一個包含 100 個 html 文件的文件夾，我正在嘗試創建一個名稱為 page1 - page100 的樹。 每個頁面都有一個超鏈接，可以打開另一個頁面。 我試圖讓它到程序讀取根節點的位置（page1.html）讀取其超鏈接並根據這些鏈接創建子節點，然后對節點的 rest 重復此操作，直到樹完成。 go 使用超鏈接的最佳方法是什么？ 到目前為止，這是我的代碼。

    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()

Answer 1

您可以使用 os.walk 遍歷所有子目錄：

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

請詢問是否有不清楚的地方。