Question

所以我使用BeautifulSoup构建了一个网络刮板来抓取Craigslist页面上的每个广告。这是我到目前为止所得到的：

import requests
from bs4 import BeautifulSoup, SoupStrainer
import bs4

page = "http://miami.craigslist.org/search/roo?query=brickell"
search_html = requests.get(page).text

roomSoup = BeautifulSoup(search_html, "html.parser")

ad_list = roomSoup.find_all("a", {"class":"hdrlnk"})
#print ad_list
ad_ls = [item["href"] for item in ad_list]
#print ad_ls
ad_urls = ["miami.craigslist.org" + ad for ad in ad_ls]
#print ad_urls 
url_str = [str(unicode) for unicode in ad_urls]

# What's in url_str?
for url in url_str:
    print url

当我跑步时，我得到：

miami.craigslist.org/mdc/roo/4870912192.html miami.craigslist.org/mdc/roo/4858122981.html miami.craigslist.org/mdc/roo/4870665175.html miami.craigslist.org/mdc/roo/4857247075.html miami.craigslist.org/mdc/roo/4870540048.html ...

这正是我想要的：一个包含页面上每个广告的网址的列表。

我的下一步是从每个页面中提取一些内容;因此构建另一个BeautifulSoup对象。但是我被迫做空了：

for url in url_str:
    ad_html = requests.get(str(url)).text

在这里，我们终于得到了一个问题：这个错误究竟是什么？我唯一能理解的是最后两行：

 Traceback (most recent call last):   File "webscraping.py", line 24,
 in <module>
     ad_html = requests.get(str(url)).text   File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/requests/api.py",
 line 65, in get
     return request('get', url, **kwargs)   File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/requests/api.py",
 line 49, in request
     response = session.request(method=method, url=url, **kwargs)   File
 "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/requests/sessions.py",
 line 447, in request
     prep = self.prepare_request(req)   File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/requests/sessions.py",
 line 378, in prepare_request
     hooks=merge_hooks(request.hooks, self.hooks),   File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/requests/models.py",
 line 303, in prepare
     self.prepare_url(url, params)   File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/requests/models.py",
 line 360, in prepare_url
     "Perhaps you meant http://{0}?".format(url)) requests.exceptions.MissingSchema: Invalid URL
 u'miami.craigslist.org/mdc/roo/4870912192.html': No schema supplied.
 Perhaps you meant http://miami.craigslist.org/mdc/roo/4870912192.html?

看起来问题是我的所有链接都以u＆＃39;开头，所以requests.get（）不起作用。这就是为什么你看到我几乎试图用str（）强制所有的URL成为常规字符串。不管我做什么，我都会遇到这个错误。还有别的东西我不见了吗？我完全误解了我的问题吗？

提前多多谢谢！

Answer 1

看起来你误解了这个问题

消息：

 u'miami.craigslist.org/mdc/roo/4870912192.html': No schema supplied.
 Perhaps you meant http://miami.craigslist.org/mdc/roo/4870912192.html?

表示在网址

之前缺少http://（架构）

所以更换

ad_urls = ["miami.craigslist.org" + ad for ad in ad_ls]

通过

ad_urls = ["http://miami.craigslist.org" + ad for ad in ad_ls]

应该做的工作

BeautifulSoup：find_all（）和unicode的问题？

1 个答案: