Python之使用来自 os.listdir 的超链接创建 html 页面

我有一个脚本可以在 USB 驱动器上创建一个名为“视频”的文件夹，将 6,500 个 WMV 文件移动到“视频”文件夹中。然后假设创建一个 HTML 页面，其中包含指向每个文件的超链接。这是我当前损坏的示例。我试图让它抓取视频目录并创建一个仅包含指向 USB 驱动器上本地文件的超链接的 HTML 页面。

#!/usr/bin/python 
import os.path 
import os 
import shutil 
import re 
 
# Create the videos directory in the current location 
# If the directory exists ignore it 
def createDirectory(): 
    directory = "videos" 
    if not os.path.isdir("./" + directory + "/"): 
        os.mkdir("./" + directory + "/") 
        print "Videos Folder Created." 
    else: 
        print "Video Folder Exists." 
        print "---------------------" 
 
# Move all the files in the root directory with the .wmv extension 
# to the videos folder 
def moveVideos(): 
    for file in os.listdir("."): 
        if os.path.splitext(file)[1] == ".wmv": 
            print "Moving:", file 
            shutil.move(file, os.path.join("videos", file)) 
 
def createHTML(): 
    videoDirectory = os.listdir("videos") 
    f = open("videos.html", "w") 
    f.writelines(videoDirectory) 
    r = re.compile(r"(\\[^ ]+)") 
    print r.sub(r'<a href="\1">\1</a>', videoDirectory) 
 
createDirectory() 
moveVideos() 
createHTML() 

请您参考如下方法:

import cgi 
 
def is_video_file(filename): 
  return filename.endswith(".wmv") # customize however you like 
 
def createHTML(): 
  videoDirectory = os.listdir("videos") 
  with open("videos.html", "w") as f: 
    f.write("<html><body><ul>\n") 
    for filename in videoDirectory: 
      if is_video_file(filename): 
        f.write('<li><a href="%s">%s</a></li>\n' % 
                (cgi.escape(filename, True), cgi.escape(filename))) 
    f.write("</ul></body></html>\n")