Pythonla çok samimi olmasakta bazen işimi çok kolaylaştırdığını(150 yi aşkın sayfayı belli bir veri için kontrol etmek gibi) inkar edemem. Basit bir kod belki ama kullanışlı olduğu kesin.
#Page.py
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import urllib
import urllib2
POST = 1
GET = 2
def retrieve(url,values,method):
----------->try:
----------------------->data=urllib.urlencode(values)
----------------------->if method == POST:
---------------------------------->request=urllib2.Request(url,data)
---------------------------------->response=urllib2.urlopen(request)
---------------------->elif method == GET:
---------------------------------->get_url=url+'?'+data
---------------------------------->response=urllib2.urlopen(get_url)
---------------------->content=response.read()
---------------------->return content
---------->except urllib2.HTTPError, e:
---------------------->print 'Gecersiz istek!!!'
---------------------->print 'Hata: ', e.code
---------->except urllib2.URLError, e:
---------------------->print 'Servera baglanti yok!!'
---------------------->print 'Sebep: ', e.reason
---------->print 'Opppsss...Fonksiyon basarisiz oldu.'
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Bu modulün kullanımına örnek olarak ta şöyle bir kod yazılabilir.
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#-*- coding: utf-8 -*- #kod içerisinde türkçe karakter kullanabilmek için
import Page
print "Trying...."
for i in range(1,179):
----------->print "%s " % i,
----------->values={'sektorid':'39','ilce_id':'','page':str(i)}
----------->content=Page.retrieve('http://www.webrehberi.net/istanbul_rehberi/alt_sektor.php',values,Page.GET)
----------->if (">ERDEK<" in content) or (">DURSUNBEY<" in content):
----------------------->print ""
----------------------->print "Found at %s. page" % i
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Herzaman ki gibi kodlara “Dosyalarım” linkinden ulaşılabilir.
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