BeautifulSoup有错误的回应

Question

我正试着用BS弄湿我的脚.我试图通过文档工作,但在我遇到问题的第一步.

这是我的代码:

from bs4 import BeautifulSoup
soup = BeautifulSoup('https://api.flickr.com/services/rest/?method=flickr.photos.search&api_key=5....1b&per_page=250&accuracy=1&has_geo=1&extras=geo,tags,views,description')

print(soup.prettify())

这是我得到的回应:

Warning (from warnings module):
File "/Library/Frameworks/Python.framework/Versions/3.3/lib/python3.3/site-packages/bs4/__init__.py", line 189
'"%s" looks like a URL. Beautiful Soup is not an HTTP client. You should probably use an     
HTTP client to get the document behind the URL, and feed that document to Beautiful Soup.' % markup)
UserWarning: "https://api.flickr.com/services/rest/?method=flickr.photos.search&api_key=5...b&per_page=250&accuracy=1&has_geo=1&extras=geo,tags,views,description" 
looks like a URL. Beautiful Soup is not an HTTP client. You should 
probably use an HTTP client to get the document behind the URL, and feed that document    
to Beautiful Soup.
https://api.flickr.com/services/rest/?method=flickr.photos.search&api;_key=5...b&per;_page=250&accuracy;=1&has;_geo=1&extras;=geo,tags,views,description

是因为我试着打电话给http**s**还是另一个问题？谢谢你的帮助!

Answer 1

您将URL作为字符串传递.相反,您需要通过urllib2或获取页面源requests:

from urllib2 import urlopen  # for Python 3: from urllib.request import urlopen
from bs4 import BeautifulSoup

URL = 'https://api.flickr.com/services/rest/?method=flickr.photos.search&api_key=5....1b&per_page=250&accuracy=1&has_geo=1&extras=geo,tags,views,description'
soup = BeautifulSoup(urlopen(URL))

请注意,您不需要调用read()结果urlopen(),BeautifulSoup允许第一个参数成为类文件对象,urlopen()返回类似文件的对象.