Python，试图通过 URL 获取文件扩展名

Question

我正在为 puush 服务制作图像采集器；但是，每当我生成随机 URL 并尝试将其验证为 .PNG 图像时，就会生成错误。我今天早些时候刚刚开始学习 Python 语言，所以你可以说我对此很陌生!

产生的错误:

Traceback (most recent call last):
  File "run.py", line 19, in <module>
    extension = guess_extension(guess_type(url))
  File "C:\Python33\lib\mimetypes.py", line 320, in guess_extension
    return _db.guess_extension(type, strict)
  File "C:\Python33\lib\mimetypes.py", line 189, in guess_extension
    extensions = self.guess_all_extensions(type, strict)
  File "C:\Python33\lib\mimetypes.py", line 168, in guess_all_extensions
    type = type.lower()
AttributeError: 'tuple' object has no attribute 'lower'

运行的代码:

#!/usr/bin/env python
import sys
import urllib
from mimetypes import guess_type, guess_extension
from random import choice

randoms = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z',
           'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I',' J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z',
           '0', '1', '2', '3', '4', '5', '6', '7', '8', '9'];

downloads = 1;

#the number of files we want to download
target = int(sys.argv[1]);

while downloads <= target:
    string = choice(randoms)+choice(randoms)+choice(randoms)+choice(randoms)+choice(randoms)
    url = 'http://puu.sh/'+string

    print(str(downloads)+': '+string)

    #download
    extension = guess_extension(guess_type(url))
    print(extension)

    #urllib.request('http://puu.sh/'+string, string+'.png')
    downloads += 1

关于此错误试图告诉我什么的任何想法？谢谢。

Answer 1

guess_type()返回元组 (type,encoding)，而 guess_extension() 接受单个参数 type。

线

extension = guess_extension(guess_type(url))

调用 guess_type，然后将其返回值(元组)传递给 guess_extension。您应该只传递元组的第一个元素 (type)

extension = guess_extension(guess_type(url)[0])