Https Form Page

[Joshua Ginsberg]

Try:

import urllib2, urllib
formdata = urllib.urlencode({'a':1,'b':2})
request = urllib2.Request(url)
request.add_data(formdata)
opener = urllib2.build_opener(urllib2.HTTPSHandler())
urllib2.install_opener(opener)
fs = urllib2.urlopen(request)
https_response = fs.read()
https_headers = fs.info().headers
https_mimetype = fs.info().type

-jag

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Joshua Ginsberg -- joshg at brainstorminternet.net
Brainstorm Internet Network Operations
970-247-1442 x131
On Apr 8, 2005, at 2:13 PM, Hasan D wrote:

> I'm new on this httplib and urllib. Actually I dont know what should i 
> use.
>
> I want to fill the form in a "https" page , and return the result . I
> write a test code but always gives errors. I cant find any good
> example about this on the net. What should I do about this ?
>
>
> import urlparse,urllib,httplib,string,htmllib,formatter
>
> #port="443"
> target="https://www.abc.com/"
> params = urllib.urlencode({'spam': 1, 'eggs': 2, 'bacon': 0})
> http=httplib.HTTP("https://www.abc.com/",443)
>
> data='Name=x&Adress=x'
> headers = {"Content-type":
> "application/x-www-form-urlencoded","Accept": "text/plain"}
>
> print "Sending Data On "+target+"...\n"
> http.putrequest("POST",target+"/xxx.asp?q=7&b=11",params)
>
> response = http.getresponse()
> print response.status, response.reason
>
> http.send(data)
>
> code,msg,headers = http.getreply()
>
> print "HTTP Code : ",str(code)
> print "HTTP Connection : ",msg
> print "HTTP headers : \n",headers,"\n"
>
> HTML=http.getfile().read()
> MyParser=htmllib.HTMLParser(formatter.NullFormatter())
> MyParser.feed(HTML)
> # Print all the anchors from the results page
> print MyParser.anchorlist
> -- 
> http://mail.python.org/mailman/listinfo/python-list
>