我一直在尝试创建一个应用程序,使用电报机器人,快速服务器和反应应用程序。因此,我需要从telegram-bot创建一个POST请求来表达,而express将POST数据发送到一个websocket连接:
const express = require("express");
const app = express();
const expressWs = require("express-ws")(app);
// handles bot request
app.post("/request", (req, res) => {
playlist.push(req.body.url);
res.status(200).send({ message: "video is added to playlist" });
});
// after handling requst data must go here and send ws message to client side
app.ws("/echo", (ws, req) => {
ws.on("message", msg => {
ws.send(`msg is = ${msg}`);
});
});我做得对吗?如果是这样的话,在处理app.post路由的请求后如何从调用ws.send?
发布于 2019-10-22 00:06:29
根据我从你的问题中得到的理解,这是你的代码的一个更新版本,它完全符合你的要求。
我用ws替换了express-ws包,因为这对您的用例来说已经足够了。
express服务器在端口8080上运行,而websocket服务器在端口8081上运行,因为它们是不同的协议,不会在同一端口上运行(您可以使其工作,但我不推荐使用See this question
const express = require("express");
const Websocket = require('ws');
const app = express();
app.use(express.json());
app.use(express.urlencoded({ extended: false }));
const wss = new Websocket.Server({ port: 8081 });
wss.on('connection', (ws) => {
console.log('One client connected');
ws.on("message", msg => {
ws.send(`msg is = ${msg}`);
});
})
// handles bot request
app.post("/request", (req, res) => {
// Broadcast URL to connected ws clients
wss.clients.forEach((client) => {
// Check that connect are open and still alive to avoid socket error
if (client.readyState === Websocket.OPEN) {
client.send(url);
}
});
res.status(200).send({ message: "video is added to playlist" });
});
app.listen(8080, () => {
console.log('Express listening on 8080');
console.log('Websocket on 8081');
});通过curl和curl -d 'url=https://example.com/examplesong' localhost:8080/request进行了测试,我有一个连接到ws://localhost:8081的客户端,一切看起来都很好。
https://stackoverflow.com/questions/58483487
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