Domanda

Data una stringa di parole chiave, come "best practice Python", vorrei ottenere le prime 10 domande di Stack Overflow che contengono tali parole chiave, ordinate per pertinenza (?), da uno script Python. Il mio obiettivo è quello di finire con un elenco di tuple (titolo, URL).

Come posso farlo? Considereresti invece di interrogare Google? (Come lo faresti da Python?)

È stato utile?

Soluzione

>>> 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?')]

La conversione in una funzione dovrebbe essere banale.

MODIFICA : Cavolo, lo farò ...

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

Altri suggerimenti

Poiché Stackoverflow ha già questa funzione, è sufficiente ottenere i contenuti della pagina dei risultati di ricerca e acquisire le informazioni necessarie. Ecco l'URL per una ricerca per pertinenza:

  

https://stackoverflow.com/search?q=python+best+practices&sort=relevance

Se visualizzi Sorgente, vedrai che le informazioni necessarie per ogni domanda sono su una linea come questa:

<h3><a href="/questions/5119/what-are-the-best-rss-feeds-for-programmersdevelopers#5150" class="answer-title">What are the best RSS feeds for programmers/developers?</a></h3>

Quindi dovresti essere in grado di ottenere i primi dieci facendo una ricerca regex per una stringa di quel modulo.

Suggerisci di aggiungere un'API REST a SO. http://stackoverflow.uservoice.com/

È possibile schermare il codice HTML restituito da una richiesta HTTP valida. Ma ciò comporterebbe un cattivo karma e la perdita della capacità di godersi una buona notte di sonno.

Vorrei solo usare Pycurl per concatenare i termini di ricerca nella query uri.

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