我向these questions询问了有关Sanso的Websocket。这是跟进。
以下内容使我可以向Websocket客户端广播消息:
from sanic import Sanic, response
from sanic.websocket import WebSocketProtocol
import asyncio
import time
app = Sanic()
@app.websocket('/feed')
async def feed(request, ws):
while True:
now = time.time()
data = f'message: {now}'
print(data)
await ws.send(data)
await asyncio.sleep(1)
@app.route('/')
async def handle_request(request):
return response.html("""
<html><head>
<script>
const ws = new WebSocket("ws://" + location.host + '/feed');
ws.onmessage = event => {
let e = event;
console.log(e.data);
let out = document.getElementById('out');
out.innerHTML += `<li><p>${e.data}</p></li>`;
}
document.querySelector('form').addEventListener('submit', (event) => {
event.preventDefault();
let message = document.querySelector("#in").value;
ws.send(message);
document.querySelector("#in").value = "";
})
</script>
</head>
<body><h1>Main</h1>
<div id="in"><form><input type="text" method="post"></form></div>
<div><ul id="out"></ul></div>
</body>
</html>
""")
app.run(host="0.0.0.0", port=8000)
我可以确认服务器既可以定期向客户端传递消息,又可以接收某种消息。
在服务器上:
[2019-08-12 22:06:27 -0500] - (sanic.access)[INFO][127.0.0.1:49028]: GET http://localhost:8000/ 200 714
message: 1565665587.4367297
message: 1565665588.4373734
message: 1565665589.4389973
message: 1565665590.440603
message: 1565665591.4414358
message: 1565665592.441888
message: 1565665593.443465
[2019-08-12 22:06:34 -0500] - (sanic.access)[INFO][127.0.0.1:49036]: GET http://localhost:8000/ 200 714
message: 1565665594.362771
message: 1565665595.3643198
message: 1565665596.3655813
message: 1565665597.3671694
在客户端上:
但是,由于以下几个原因,这对我来说没有意义:
event.preventDefault()
被使用了,我仍然看到表单提交上有完整的页面刷新。但是,我真正想要的是双向通信。因此,除了ws.send(data)
外,我还将在方法中转到new = await ws.recv()
并捕获输入。
@app.websocket('/feed')
async def feed(request, ws):
while True:
now = time.time()
data = f'message: {now}'
print(data)
await ws.send(data)
new = await ws.recv() # this is the only change
print(new)
await asyncio.sleep(1)
但是现在,我不再继续向客户端发送数据。相反,该进程挂起,即使我的客户端没有发送任何东西,也要等待我的ws.recv()
。
[2019-08-12 22:13:52 -0500] [12920] [INFO] Starting worker [12920]
[2019-08-12 22:13:56 -0500] - (sanic.access)[INFO][127.0.0.1:49086]: GET http://localhost:8000/ 200 714
message: 1565666037.0688074
[2019-08-12 22:14:03 -0500] - (sanic.access)[INFO][127.0.0.1:49090]: GET http://localhost:8000/ 200 714
在我再次提交表单之前,不会再发送message
。
我如何进行双向通信,将数据传递回/feed
端点,但是连续不断地等待而不需等待?
答案 0 :(得分:2)
您需要将消费者和生产者分开,下面是一个示例:
from sanic import Sanic, response
import asyncio
import time
app = Sanic()
async def _consumer_handler(ws):
print('consumer_handler')
while True:
message = await ws.recv()
print('message arrived', message)
async def _producer_handler(ws):
print('producer_handler')
while True:
now = time.time()
data = f'message sent: {now}'
print(data)
await ws.send(data)
await asyncio.sleep(1)
@app.websocket('/feed')
async def feed(request, ws):
consumer_task = asyncio.ensure_future(
_consumer_handler(ws))
producer_task = asyncio.ensure_future(
_producer_handler(ws))
done, pending = await asyncio.wait(
[consumer_task, producer_task],
return_when=asyncio.FIRST_COMPLETED,
)
for task in pending:
task.cancel()
@app.route('/')
async def handle_request(request):
return response.html("""
<html><head>
</head>
<body><h1>Main</h1>
<div><form><input id="in" type="text" method="post"></form></div>
<div><ul id="out"></ul></div>
</body>
<script>
const ws = new WebSocket("ws://" + location.host + '/feed');
ws.onmessage = event => {
let e = event;
console.log(e.data);
let out = document.getElementById('out');
out.innerHTML += `<li><p>${e.data}</p></li>`;
}
document.querySelector('form').addEventListener('submit', (event) => {
event.preventDefault();
let message = document.querySelector("#in").value;
ws.send(message);
document.querySelector("#in").value = "";
})
</script>
</html>
""")
app.run(host="0.0.0.0", port=8000)
要在消费者和生产者之间进行通信,您需要使用pubsub或定期检查新消息(并且您需要将消息存储在某个地方,例如可以使用redis)。
一些有用的链接: