>>> import httplib
>>> x = httplib.HTTPConnection('localhost', 8080)
>>> x.connect()
>>> x.request('GET','/camera/store?fn=aaa&ts='+str.encode('2015-06-15T14:45:21.982600+00:00','ascii')+'&cam=ddd')
>>> y=x.getresponse()
>>> z=y.read()
>>> z
'error: Invalid format: "2015-06-15T14:45:21.982600 00:00" is malformed at " 00:00"'
And the system show me this error. As i want to encode this format to this: 2015-06-15T14%3A45%3A21.982600%2B00%3A00
URLs cannot contain spaces. URL encoding normally replaces a space with a plus (+) sign or with %20.
URL encoding converts characters into a format that can be transmitted over the Internet. - w3Schools. So, "/" is actually a seperator, but "%2f" becomes an ordinary character that simply represents "/" character in element of your url.
replace('%20+', '') will replace '%20+' with empty string.
>>> import urllib
>>> f = { 'fn' : 'aaa', 'ts' : "2015-06-15T14:45:21.982600+00:00"}
>>> urllib.urlencode(f)
from:
How to urlencode a querystring in Python?
url = "http://example.com?p=" + urllib.quote(query)
it works with this!
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