我怎样才能从一个脚本堆栈溢出的问题进行搜索？

StackOverflow https://stackoverflow.com/questions/196755

题

由于关键字的字符串，如“巨蟒最佳实践”，我想获得包含的关键字，通过关联（？）排序，从Python脚本说出的第10点堆栈溢出的问题。我的目标是与元组的列表（标题，URL）来结束。

我怎样才能做到这一点？你会考虑查询谷歌呢？ （你会如何用Python做呢？）

有帮助吗？

解决方案 

>>> from urllib import urlencode
>>> params = urlencode({'q': 'python best practices', 'sort': 'relevance'})
>>> params
'q=python+best+practices&sort=relevance'
>>> from urllib2 import urlopen
>>> html = urlopen("http://stackoverflow.com/search?%s" % params).read()
>>> import re
>>> links = re.findall(r'<h3><a href="([^"]*)" class="answer-title">([^<]*)</a></h3>', html)
>>> links
[('/questions/5119/what-are-the-best-rss-feeds-for-programmersdevelopers#5150', 'What are the best RSS feeds for programmers/developers?'), ('/questions/3088/best-ways-to-teach-a-beginner-to-program#13185', 'Best ways to teach a beginner to program?'), ('/questions/13678/textual-versus-graphical-programming-languages#13886', 'Textual versus Graphical Programming Languages'), ('/questions/58968/what-defines-pythonian-or-pythonic#59877', 'What defines &#8220;pythonian&#8221; or &#8220;pythonic&#8221;?'), ('/questions/592/cxoracle-how-do-i-access-oracle-from-python#62392', 'cx_Oracle - How do I access Oracle from Python? '), ('/questions/7170/recommendation-for-straight-forward-python-frameworks#83608', 'Recommendation for straight-forward python frameworks'), ('/questions/100732/why-is-if-not-someobj-better-than-if-someobj-none-in-python#100903', 'Why is if not someobj: better than if someobj == None: in Python?'), ('/questions/132734/presentations-on-switching-from-perl-to-python#134006', 'Presentations on switching from Perl to Python'), ('/questions/136977/after-c-python-or-java#138442', 'After C++ - Python or Java?')]
>>> from urlparse import urljoin
>>> links = [(urljoin('http://stackoverflow.com/', url), title) for url,title in links]
>>> links
[('http://stackoverflow.com/questions/5119/what-are-the-best-rss-feeds-for-programmersdevelopers#5150', 'What are the best RSS feeds for programmers/developers?'), ('http://stackoverflow.com/questions/3088/best-ways-to-teach-a-beginner-to-program#13185', 'Best ways to teach a beginner to program?'), ('http://stackoverflow.com/questions/13678/textual-versus-graphical-programming-languages#13886', 'Textual versus Graphical Programming Languages'), ('http://stackoverflow.com/questions/58968/what-defines-pythonian-or-pythonic#59877', 'What defines &#8220;pythonian&#8221; or &#8220;pythonic&#8221;?'), ('http://stackoverflow.com/questions/592/cxoracle-how-do-i-access-oracle-from-python#62392', 'cx_Oracle - How do I access Oracle from Python? '), ('http://stackoverflow.com/questions/7170/recommendation-for-straight-forward-python-frameworks#83608', 'Recommendation for straight-forward python frameworks'), ('http://stackoverflow.com/questions/100732/why-is-if-not-someobj-better-than-if-someobj-none-in-python#100903', 'Why is if not someobj: better than if someobj == None: in Python?'), ('http://stackoverflow.com/questions/132734/presentations-on-switching-from-perl-to-python#134006', 'Presentations on switching from Perl to Python'), ('http://stackoverflow.com/questions/136977/after-c-python-or-java#138442', 'After C++ - Python or Java?')]

此转换为一个函数应该是微不足道的。

修改：哎呀，我会做到这一点... 

def get_stackoverflow(query):
    import urllib, urllib2, re, urlparse
    params = urllib.urlencode({'q': query, 'sort': 'relevance'})
    html = urllib2.urlopen("http://stackoverflow.com/search?%s" % params).read()
    links = re.findall(r'<h3><a href="([^"]*)" class="answer-title">([^<]*)</a></h3>', html)
    links = [(urlparse.urljoin('http://stackoverflow.com/', url), title) for url,title in links]

    return links

其他提示

您可以筛选从一个有效的HTTP请求刮返回的HTML。但是，这将导致恶业，并享有良好的夜间睡眠的能力丧失。

我只想用Pycurl来连接搜索项到查询URI。

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