Python扭曲的JSON解码

Question

我已经使用python扭曲库开发了以下代码：

class Cache(protocol.Protocol):
    def __init__(self, factory):
        self.factory = factory

    def dataReceived(self, data):
        request = json.loads(data)
        self.factory.handle[request['command']](**request)
        self.transport.write(data)

class CacheFactory(protocol.Factory):
    def buildProtocol(self, addr):
        return Cache(self)
    def handle_get(self, **kwargs):
        print 'get\n', kwargs
    def handle_set(self, **kwargs):
        print 'set\n', kwargs
    def handle_delete(self, **kwargs):
        print 'delete\n', kwargs
    handle = {
        'get': handle_get,
        'set': handle_set,
        'delete': handle_delete,
    }

reactor.listenTCP(int(sys.argv[1]), CacheFactory())
reactor.run()

我使用telnet运行客户端连接：

Trying 127.0.0.1...
Connected to localhost.
Escape character is '^]'.
{"command": "set", "value": 1234567890}
Connection closed by foreign host.

引发异常：

Traceback (most recent call last):
  File "/usr/lib/python2.7/dist-packages/twisted/python/log.py", line 84, in callWithLogger
    return callWithContext({"system": lp}, func, *args, **kw)
  File "/usr/lib/python2.7/dist-packages/twisted/python/log.py", line 69, in callWithContext
    return context.call({ILogContext: newCtx}, func, *args, **kw)
  File "/usr/lib/python2.7/dist-packages/twisted/python/context.py", line 118, in callWithContext
    return self.currentContext().callWithContext(ctx, func, *args, **kw)
  File "/usr/lib/python2.7/dist-packages/twisted/python/context.py", line 81, in callWithContext
    return func(*args,**kw)
--- <exception caught here> ---
  File "/usr/lib/python2.7/dist-packages/twisted/internet/selectreactor.py", line 146, in _doReadOrWrite
    why = getattr(selectable, method)()
  File "/usr/lib/python2.7/dist-packages/twisted/internet/tcp.py", line 460, in doRead
    rval = self.protocol.dataReceived(data)
  File "./server.py", line 18, in dataReceived
    self.factory.handle[request['command']](**request)
exceptions.TypeError: handle_set() takes exactly 1 argument (0 given)

我不明白。 self.factory.handle[request['command']](**request)行可能有问题，但我认为这是正确的-它隐式地传递了self参数（毕竟是方法）并显式地解压缩了请求参数。 异常消息说该函数接受1个参数，这是一个谎言:)，因为它接受2个参数： self, **kwargs 。 而且我传递了2个参数也不是正确的，因为我传递了2个参数。

有人可以帮助我发现问题吗？

---

如果有帮助，则将json请求解码为：

{u'command': u'set', u'value': 1234567890}

Answer 1

到目前为止， handle_*方法是实例方法，而handle字典指向未绑定的方法。 也就是说， self不会被隐式传递。 尝试以下方法：

class CacheFactory(protocol.Factory):
    def buildProtocol(self, addr):
        return Cache(self)
    def handle_get(self, **kwargs):
        print 'get\n', kwargs
    def handle_set(self, **kwargs):
        print 'set\n', kwargs
    def handle_delete(self, **kwargs):
        print 'delete\n', kwargs
    def __init__(self, *args, **kwargs):
        protocol.Factory.__init__(self, *args, **kwargs)
        self.handle = {
            'get': self.handle_get,
            'set': self.handle_set,
            'delete': self.handle_delete,
        }

另外，您可以保持handle不变并执行以下操作：

    def dataReceived(self, data):
        request = json.loads(data)
        self.factory.handle[request['command']](self.factory, **request)
        self.transport.write(data)

或者，你可以采取这种做法，那么你并不需要一个handle字典两种方式：

    def dataReceived(self, data):
        request = json.loads(data)
        getattr(self.factory, "handle_%s" % (request['command'],))(**request)
        self.transport.write(data)

还要注意，由于数据包可能会被任意拆分，因此dataReceived （现在）是不安全的，也就是说，您可能不会一枪就收到整个json消息。