Question

在Python中，我正在为Websocket客户端使用“ websockets”库。

import asyncio
import websockets

async def init_sma_ws():
    uri = "wss://echo.websocket.org/"
    async with websockets.connect(uri) as websocket:
        name = input("What's your name? ")

        await websocket.send('name')
        greeting = await websocket.recv()

问题是，一旦收到响应，客户端Websocket连接就会断开。我希望连接保持打开状态，以便以后可以发送和接收消息。

我需要进行哪些更改以保持websocket打开并以后能够发送和接收消息？

Answer 1

我认为您的网络套接字由于recv()之后退出上下文管理器而被断开了连接。这样的代码可以完美地工作：

import asyncio
import websockets


async def init_sma_ws():
    uri = "wss://echo.websocket.org/"
    async with websockets.connect(uri) as websocket:
        while True:
            name = input("What's your name? ")
            if name == 'exit':
                break

            await websocket.send(name)
            print('Response:', await websocket.recv())


asyncio.run(init_sma_ws())

Python Websockets客户端保持连接打开

1 个答案: