"urllib2", \"httplib","cgilib")#将地址解析成组件print"用Google搜索python时地址栏中URL的解析结果"parsedTuple=urlparse.urlparse("http://www.google.com/search?hl=en&q=python&btnG=Google+Search")printparsedTuple#将组件反解析成URLprint"\反解析python文档页面的URL"unparsedURL=urlparse...
下面是一个简单的Python程序，实现了对某个网站的递归抓取：pythonimport requestsfrom bs4 import BeautifulSoupimport revisited_urls = set()def fetch_url(url): response = requests.get(url) if response.status_code == 200: visited_urls.add(url) return response.text else: return No...
>>> urlparse('//www.cwi.nl:80/%7Eguido/Python.html') ParseResult(scheme='', netloc='www.cwi.nl:80', path='/%7Eguido/Python.html', params='', query='', fragment='') >>> urlparse('www.cwi.nl/%7Eguido/Python.html') ParseResult(scheme='', netloc='', path='www.cwi.nl/%...
like Gecko) Chrome/58.0.3029.110 Safari/537.3'} r = requests.get(url, headers=headers) soup = BeautifulSoup(r.text,'html.parser') links =[] for link in soup.find_all('a'): links.append
/usr/bin/python#-*- coding: UTF-8 -*-frompyqueryimportPyQuery as pq classUrlParser():def__init__(self): self.urls = []deffeed(self,data): d = pq(data)ifd.find('a'):#关于下面一行,我用d('a').attr('href')只能得到第一个URL,暂时只会用map,不知道有没有别的够pythonic的代码url...
html = response.text #分析页面内容，提取需要的数据信息 soup = BeautifulSoup(html,'html.parser') data = extract_data(soup) #存储数据到指定位置中 save_data(data) #获取该页面上包含的所有链接，并加入到待遍历队列中 links = extract_links(soup) for link in links: url...
Python3 requests parsel threading queue argparse sys 其实简单点只需要前面三个库就好了,只是强迫症就想搞正规点。 大概思路 输入关键字,分析url ,以及翻页。 模拟爬取,提取出单页所需url。 实现多页爬取。 多线程爬取+可控参数 1. 分析url
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pythonimport requestsfrom bs4 import BeautifulSoupdef get_links(url): links =[] r = requests.get(url) soup = BeautifulSoup(r.text,'html.parser') for link in soup.find_all('a'): href = link.get('href') if href.b1d1e10add929d03ff21980c48ec0769('http'): links...