httplib won't accept URL with parameters (???)

[pythonhda]

See the code changes below...


On Mon, 06 Jan 2003 23:28:09 +0100
Phillip <ritschratsch at gmx.de> wrote:

> I've got some serious httplib Problem. Here's the relevant code:
> 
> USER_AGENT = "Navigator" # We have to give something, don't we?
> 
> class ServerGet:
> 
> 	def __init__(self, host):
> 		self.host = host
> 	def fetch(self, path):
> 		http = httplib.HTTP(self.host)
> 
> 		#write header
> 		http.putrequest("GET", path)
> 		http.putheader("User-Agent", USER_AGENT)
> 		http.putheader("Host", self.host)
> 		http.putheader("Accept", "*/*")
> 		http.endheaders()
> 
> 		# get response
> 		errcode, errmsg, headers = http.getreply()
> 
> 
> 		file = http.getfile()
 		data = file.read() ####
		http.close()       ####
		return data        ####


> 
> CompareServer = ServerGet("www.server.com")
> ComparePage = 
> CompareServer.fetch("/grab/database/get-page/link=ddb_benutzt_a/00b-97rt692-8ab6561?view=fixed-thing&field-stuff-type=used&thing-number=3897211920&field-status=open&size=25&rank=+parameter")
> # long URL, I know....
> print ComparePage
> 
> The Problem is that CompareServer.fetch won't return anything(!!) I 
> presume it has something to do with the fact that the URL that is passed 
> on has a tail of parameters trailing and that httplib won't accept this.
> Does anyone know how I can get this to work? Is it a 'GET' problem?
> Note that httplib work perfect with the usual URLs that only have 
> slashes in them.
> What's the big fuss? Why doesn't ServerGet return something with that 
> URL? Any clue?
> Thanks for all help in advance.
> 
> Phillip
>