BeautifulSoup:find_all() 和 unicode 有问题吗?

BeautifulSoup: An issue with find_all() and unicode?

所以我正在使用 BeautifulSoup 构建一个网络爬虫来抓取 Craigslist 页面上的每个广告。到目前为止,这是我得到的:

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 运行 这个时,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:包含页面上每个广告的 URL 的列表。

我的下一步是从每个页面中提取一些东西;因此构建另一个 BeautifulSoup 对象。但是我突然停下来了:

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

我们终于到了我的问题:这个错误到底是什么?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?

问题似乎是我的所有链接都以 u' 开头,因此 requests.get() 不起作用。这就是为什么你看到我几乎试图用 str() 将所有 URL 强制转换为常规字符串。但是,无论我做什么,我都会收到此错误。还有什么我想念的吗?我完全误解了我的问题吗?

在此先致谢!

看来你误解了问题

留言:

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

表示它在 url

之前缺少 http://(架构)

所以替换

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

来自

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

应该做这份工作