我必须用 Python 编写一个网络爬虫。我不知道如何解析页面并从 HTML 中提取 URL。我应该去哪里学习编写这样的程序？
换句话说，是否有一个简单的 python 程序可以用作通用网络爬虫的模板？理想情况下，它应该使用相对简单易用的模块，并且应该包含大量注释来描述每一行代码的作用。
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62188 次
5 回答
21
看下面的示例代码。该脚本提取网页（此处为 Python 主页）的 html 代码并提取该页面中的所有链接。希望这可以帮助。
#!/usr/bin/env python
import requests
from bs4 import BeautifulSoup
url = "http://www.python.org"
response = requests.get(url)
# parse html
page = str(BeautifulSoup(response.content))
def getURL(page):
"""
:param page: html of web page (here: Python home page)
:return: urls in that page
"""
start_link = page.find("a href")
if start_link == -1:
return None, 0
start_quote = page.find('"', start_link)
end_quote = page.find('"', start_quote + 1)
url = page[start_quote + 1: end_quote]
return url, end_quote
while True:
url, n = getURL(page)
page = page[n:]
if url:
print(url)
else:
break
输出：
/
#left-hand-navigation
#content-body
/search
/about/
/news/
/doc/
/download/
/getit/
/community/
/psf/
http://docs.python.org/devguide/
/about/help/
http://pypi.python.org/pypi
/download/releases/2.7.3/
http://docs.python.org/2/
/ftp/python/2.7.3/python-2.7.3.msi
/ftp/python/2.7.3/Python-2.7.3.tar.bz2
/download/releases/3.3.0/
http://docs.python.org/3/
/ftp/python/3.3.0/python-3.3.0.msi
/ftp/python/3.3.0/Python-3.3.0.tar.bz2
/community/jobs/
/community/merchandise/
/psf/donations/
http://wiki.python.org/moin/Languages
http://wiki.python.org/moin/Languages
http://www.google.com/calendar/ical/b6v58qvojllt0i6ql654r1vh00%40group.calendar.google.com/public/basic.ics
http://www.google.com/calendar/ical/j7gov1cmnqr9tvg14k621j7t5c%40group.calendar.google.com/public/basic.ics
http://pycon.org/#calendar
http://www.google.com/calendar/ical/3haig2m9msslkpf2tn1h56nn9g%40group.calendar.google.com/public/basic.ics
http://pycon.org/#calendar
http://www.psfmember.org
...
于 2013-03-20T07:24:34.443 回答
17
您可以使用BeautifulSoup，正如许多人所说的那样。它可以解析 HTML、XML 等。要查看它的一些功能，请参见此处。
例子：
import urllib2
from bs4 import BeautifulSoup
url = 'http://www.google.co.in/'
conn = urllib2.urlopen(url)
html = conn.read()
soup = BeautifulSoup(html)
links = soup.find_all('a')
for tag in links:
link = tag.get('href',None)
if link is not None:
print link
于 2013-03-20T08:08:59.273 回答
6
import sys
import re
import urllib2
import urlparse
tocrawl = set(["http://www.facebook.com/"])
crawled = set([])
keywordregex = re.compile('<meta\sname=["\']keywords["\']\scontent=["\'](.*?)["\']\s/>')
linkregex = re.compile('<a\s*href=[\'|"](.*?)[\'"].*?>')
while 1:
try:
crawling = tocrawl.pop()
print crawling
except KeyError:
raise StopIteration
url = urlparse.urlparse(crawling)
try:
response = urllib2.urlopen(crawling)
except:
continue
msg = response.read()
startPos = msg.find('<title>')
if startPos != -1:
endPos = msg.find('</title>', startPos+7)
if endPos != -1:
title = msg[startPos+7:endPos]
print title
keywordlist = keywordregex.findall(msg)
if len(keywordlist) > 0:
keywordlist = keywordlist[0]
keywordlist = keywordlist.split(", ")
print keywordlist
links = linkregex.findall(msg)
crawled.add(crawling)
for link in (links.pop(0) for _ in xrange(len(links))):
if link.startswith('/'):
link = 'http://' + url[1] + link
elif link.startswith('#'):
link = 'http://' + url[1] + url[2] + link
elif not link.startswith('http'):
link = 'http://' + url[1] + '/' + link
if link not in crawled:
tocrawl.add(link)
参考：Python Web Crawler in Less Than 50 Lines（速度慢或不再工作，不适合我）
于 2013-03-20T08:07:00.990 回答
5
您可以使用beautifulsoup。遵循文档并查看符合您要求的内容。该文档还包含有关如何提取 URL 的代码片段。
from bs4 import BeautifulSoup
soup = BeautifulSoup(html_doc)
soup.find_all('a') # Finds all hrefs from the html doc.
于 2013-03-20T07:28:27.190 回答
3
通过解析页面，查看BeautifulSoup模块。它使用简单，并允许您使用 HTML 解析页面。您可以简单地从 HTML 中提取 URLstr.find('a')
于 2013-03-20T07:27:00.457 回答