3Thiago
commented
5 years ago Hi Bro,
Did you solve this problem? I have the same problem!
return AuthServiceProxy('http://%s:%s@%s:%i]'%(rpc_username, rpc_password, rpc_ip, rpc_port)) File "/usr/local/lib/python3.6/dist-packages/bitcoinrpc/authproxy.py", line 86, in init self.__url = urlparse.urlparse(service_url) File "/usr/lib/python3.6/urllib/parse.py", line 368, in urlparse splitresult = urlsplit(url, scheme, allow_fragments) File "/usr/lib/python3.6/urllib/parse.py", line 418, in urlsplit raise ValueError("Invalid IPv6 URL") ValueError: Invalid IPv6 URL
Hello, I am running this in a Jupyter Notebook to test it.
Python - 2.7 My RPC server is on a AWS instance.
I get the following error on this call: AuthServiceProxy("http://%s:%s@XX.XXX.XX.XXX:9009"%(rpc_user, rpc_password))
/anaconda/lib/python2.7/site-packages/bitcoinrpc/authproxy.pyc in init(self, service_url, service_name, timeout, connection) 84 self.service_url = service_url 85 self.service_name = service_name ---> 86 self.__url = urlparse.urlparse(service_url) 87 if self.__url.port is None: 88 port = 80
//anaconda/lib/python2.7/urlparse.pyc in urlparse(url, scheme, allow_fragments) 141 Note that we don't break the components up in smaller bits 142 (e.g. netloc is a single string) and we don't expand % escapes.""" --> 143 tuple = urlsplit(url, scheme, allow_fragments) 144 scheme, netloc, url, query, fragment = tuple 145 if scheme in uses_params and ';' in url:
//anaconda/lib/python2.7/urlparse.pyc in urlsplit(url, scheme, allow_fragments) 189 if (('[' in netloc and ']' not in netloc) or 190 (']' in netloc and '[' not in netloc)): --> 191 raise ValueError("Invalid IPv6 URL") 192 if allow_fragments and '#' in url: 193 url, fragment = url.split('#', 1)
ValueError: Invalid IPv6 URL
How to resolve this?