使用 python 爬取网站

Question

So I am looking for a dynamic way to crawl a website and grab links from each page. I decided to experiment with Beauitfulsoup. Two questions: How do I do this more dynamically then using nested while statements searching for links. I want to get all the links from this site. But I don't want to continue to put nested while loops.

    topLevelLinks = self.getAllUniqueLinks(baseUrl)
    listOfLinks = list(topLevelLinks)       

    length = len(listOfLinks)
    count = 0       

    while(count < length):

        twoLevelLinks = self.getAllUniqueLinks(listOfLinks[count])
        twoListOfLinks = list(twoLevelLinks)
        twoCount = 0
        twoLength = len(twoListOfLinks)

        for twoLinks in twoListOfLinks:
            listOfLinks.append(twoLinks)

        count = count + 1

        while(twoCount < twoLength):
            threeLevelLinks = self.getAllUniqueLinks(twoListOfLinks[twoCount])  
            threeListOfLinks = list(threeLevelLinks)

            for threeLinks in threeListOfLinks:
                listOfLinks.append(threeLinks)

            twoCount = twoCount +1



    print '--------------------------------------------------------------------------------------'
    #remove all duplicates
    finalList = list(set(listOfLinks))  
    print finalList

My second questions is there anyway to tell if I got all the links from the site. Please forgive me, I am somewhat new to python (year or so) and I know some of my processes and logic might be childish. But I have to learn somehow. Mainly I just want to do this more dynamic then using nested while loop. Thanks in advance for any insight.

Answer 1

The problem of spidering over a web site and getting all the links is a common problem. If you Google search for "spider web site python" you can find libraries that will do this for you. Here's one I found:

http://pypi.python.org/pypi/spider.py/0.5

Even better, Google found this question already asked and answered here on StackOverflow:

Anyone know of a good Python based web crawler that I could use?

Answer 2

If using BeautifulSoup, why don't you use findAll() method?? Basically, in my crawler i do:

self.soup = BeautifulSoup(HTMLcode)
for frm in self.soup.findAll(str('frame')):
try:
    if not frm.has_key('src'):
        continue
    src = frm[str('src')]
    #rest of URL processing here
except Exception, e:
    print  'Parser <frame> tag error: ', str(e)

for the frame tag. The same goes for "img src"and "a href" tags. I like the topic though - maybe its me who has sth wrong here... edit: there is ofc a top-level instance, which saves the URLs and gets the HTMLcode from each link later...

Answer 3

To answer your question from the comment, here's an example (it's in ruby, but I don't know python, and they are similar enough for you to be able to follow along easily):

#!/usr/bin/env ruby

require 'open-uri'

hyperlinks = []
visited = []

# add all the hyperlinks from a url to the array of urls
def get_hyperlinks url
  links = []
  begin
    s = open(url).read
    s.scan(/(href|src)\w*=\w*[\",\']\S+[\",\']/) do
      link = $&.gsub(/((href|src)\w*=\w*[\",\']|[\",\'])/, '')
      link = url + link if link[0] == '/'

      # add to array if not already there
      links << link if not links =~ /url/
    end
  rescue
    puts 'Looks like we can\'t be here...'
  end
  links
end

print 'Enter a start URL: '
hyperlinks << gets.chomp
puts 'Off we go!'
count = 0
while true
  break if hyperlinks.length == 0
  link = hyperlinks.shift
  next if visited.include? link
  visited << link
  puts "Connecting to #{link}..."
  links = get_hyperlinks(link)
  puts "Found #{links.length} links on #{link}..."
  hyperlinks = links + hyperlinks
  puts "Moving on with #{hyperlinks.length} links left...\n\n"
end

sorry about the ruby, but its a better language:P and shouldn't be hard to adapt or, like i said, understand.

Answer 4

1) In Python, we do not count elements of a container and use them to index in; we just iterate over its elements, because that's what we want to do.

2) To handle multiple levels of links, we can use recursion.

def followAllLinks(self, from_where):
    for link in list(self.getAllUniqueLinks(from_where)):
        self.followAllLinks(link)

This does not handle cycles of links, but neither did the original approach. You can handle that by building a set of already-visited links as you go.

Answer 5

Use scrapy :

Scrapy is a fast high-level screen scraping and web crawling framework, used to crawl websites and extract structured data from their pages. It can be used for a wide range of purposes, from data mining to monitoring and automated testing.