jsonrpcclient 1.0.4

JSON-RPC 2.0 client library for Python 3

A JSON-RPC 2.0 client library for Python 3.

To make a remote procedure call:

1. Set the server address

2. Use request() to call the remote method

>>> import jsonrpcclient
>>> server = jsonrpcclient.Server('http://endpoint/')
>>> server.request('add', 2, 3)
--> {"jsonrpc": "2.0", "method": "add", "params": [2, 3], "id": 1}
<-- {"jsonrpc": "2.0", "result": 5, "id": 1}
5

The first argument to request() is the method name; everything else is passed as parameters. You can pass any number of positional or keyword arguments, and they will be translated into JSON-RPC.

>>> result = server.request('find', 42, name='Foo')
--> {"jsonrpc": "2.0", "method": "find", "params": [42, {"name": "Foo"}], "id": 1}
<-- {"jsonrpc": "2.0", "result": "Bar", "id": 1}
Bar

If you don’t need any data returned, use notify() instead:

>>> server.notify('go')
--> {"jsonrpc": "2.0", "method": "go"}
<-- 200 OK

Shorthand

There’s another way to send messages:

>>> server.add(2, 3, response=True)
--> {"jsonrpc": "2.0", "method": "add", "params": [2, 3], "id": 1}
<-- {"jsonrpc": "2.0", "result": 5, "id": 1}
5

The library catches the undefined add() call, and sends it as a JSON-RPC message.

response=True tells the server you’re expecting a response; without that it’s a notification.

Exceptions

You should catch RPCClientException. This will be raised in the event of connection problems, or if the server responded with a JSON-RPC error response.

try:
    server.go()
except jsonrpcclient.exceptions.RPCClientException as e:
    print(str(e))

Logging

If you don’t want to see the underlying JSON messages, increase the logging level above DEBUG:

logging.getLogger('jsonrpcclient').setLevel(logging.INFO)

If you need a server, try my jsonrpcserver library.