对指定URL获取其titile

标签：爬虫

最近在广告投放时需要找到一批强项关的人群, 现在发现了指定的一些URL可能会跟给广告相关,所以需要对每个URL 的网页内容进行解析,以便能判断URL 是否与该广告相关.

我这里使用python中的urllib或urllib2包对URL 的内容提取.方法如下:

#!/usr/bin/python
# -*- coding: utf-8 -*-
import urllib2
import re

url='http://segmentfault.com/q/1010000000124036'
html = urllib2.urlopen(url).read()

res_list = re.findall(r"<title>.*</title>", html)
for t in res_list:
	print t

python如何正确抓取网页标题 - SegmentFault

看这个帖子里说的BeautifulSoup也是很不错的, 没有试过,代码如下:

import urllib
from BeautifulSoup import BeautifulSoup
content = urllib.urlopen('http://segmentfault.com/q/1010000000124036').read()
soup = BeautifulSoup(content)
print soup.find('title')

用第一段代码在获取我的主页http://blog.csdn.net/lming_08?viewmode=list的title时,报错了:

Traceback (most recent call last):
  File "get_title.py", line 9, in <module>
    html = urllib2.urlopen(url).read()
  File "/usr/lib/python2.7/urllib2.py", line 126, in urlopen
    return _opener.open(url, data, timeout)
  File "/usr/lib/python2.7/urllib2.py", line 400, in open
    response = meth(req, response)
  File "/usr/lib/python2.7/urllib2.py", line 513, in http_response
    'http', request, response, code, msg, hdrs)
  File "/usr/lib/python2.7/urllib2.py", line 438, in error
    return self._call_chain(*args)
  File "/usr/lib/python2.7/urllib2.py", line 372, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.7/urllib2.py", line 521, in http_error_default
    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 403: Forbidden

url='http://blog.csdn.net/lming_08?viewmode=list'

user_agent = 'Mozilla/4.0 (compatible; MSIE 5.5; Windows NT)'# 将user_agent写入头信息
headers = { 'User-Agent' : user_agent }

req = urllib2.Request(url, headers = headers)
html = urllib2.urlopen(req).read()
res_list = re.findall(r"<title>.*\s?.*</title>", html)
for t in res_list:
	print t[7:-8]

lming_08技术博客
        - 博客频道 - CSDN.NET

注意这里标题里面含有\n字符,而之前正则表达式r"<title>.*</title>"是不会包含\n的,所以改为 r"<title>.*\s?.*</title>"  或者 r"<title>.*\n?.*</title>"

最新发现这获取 http://www.smzdm.com/p/664187 title时仍然会报403错误, 查了下需要更新headers

user_agent = 'Mozilla/5.0 (Windows; U; Windows NT 6.1; en-US; rv:1.9.1.6) Gecko/20091201 Firefox/3.5.6'
headers = { 'User-Agent' : user_agent }

mingliang@ubuntu:~/MyWorkSpace/Pycode/htmlparse$ python get_title.py
Mentholatum 曼秀雷敦  肌研极润保湿系列套装（洁面乳50g+化妆水100ml+乳液90ml+面膜18ml+眼膜2片+眼霜3g） 99元（199-100）_乐蜂网优惠_什么值得买

try:
    html = urllib2.urlopen(req).read()
    res_list = re.findall(r"<title>.*\s?.*</title>", html)
    for t in res_list:
	    print t[7:-8]
except urllib2.HTTPError:
    print "failed parsing web url"

文章参考于:

http://www.cnblogs.com/txw1958/archive/2012/03/12/2392067.html

http://segmentfault.com/q/1010000000124036

http://blog.csdn.net/my2010sam/article/details/17398807

对指定URL获取其titile

标签：爬虫

原文地址：http://blog.csdn.net/lming_08/article/details/44710779