如何从 Python 中的 HTML 页面中提取 URL [关闭]

Question

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，有没有一个简单的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 cmets 来描述每行代码的作用。

Answer 1

查看下面的示例代码。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。希望这会有所帮助。

#!/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

...

Answer 2

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，请参阅here。

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

Answer 3

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）

Answer 4

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 sn-ps。

from bs4 import BeautifulSoup
soup = BeautifulSoup(html_doc)

soup.find_all('a') # Finds all hrefs from the html doc.

Answer 5

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。您可以通过 str.find('a')

从 HTML 中提取 URL

Don't use regular expressions for parsing HTML