我正在尝试使用Twitter Stream API创建流。
我有一个开始按钮,用于开始直播。
我有一个停止按钮以停止流播放。
我使用Web套接字,由此单击“开始”按钮时,我开始在服务器端进行流式传输。
当单击“停止”按钮时,我将销毁流_stream.abort()
(我尝试过_stream.destroy()
-同样的错误)
index.js(服务器端)
io.on('connection', function(socket) {
// Socket Stream Button
const _stream = {};
// Destroy on stopBtn is clicked
socket.on('myapp:search', function(term) {
console.log('received a new search:', term);
_stream.abort();
});
// Listen to Start Button being clicked, value is emitted, start twitter stream
socket.on('startStreaming', function() {
console.log("arguments::: ",arguments['0'])
twitterStream(socket, _stream, arguments['0'], function(_stream) {
_stream.on('data', function(tweet) {
console.log(tweet.text + "Streaming");
// socket.emit('tweet', tweet.text);
});
_stream.on('error', function(error) {
console.log("erorr:: " + error);
throw error;
});
});
});
}
function twitterStream(socket, _stream, term, callback) {
// return new Promise((resolve, reject) => {
console.log("TERM _" +term);
client.stream('statuses/filter', { track: term }, function(stream) {
_stream = stream;
callback(_stream);
});
// });
}
在这里,我只需单击“开始/停止”按钮即可触发发射事件。
Script.js(前端)
function sendForm() {
$("#startBtn").on('click', function() {
//input(type="text" name="searchTerm" placeholder="#hastag" required)
let searchTerm = $("#searchedTerm").val();
console.log(searchTerm);
socket.emit('startStreaming', searchTerm, function(data) {
console.log("stared");
});
});
$("#stopBtn").on('click', function() {
console.log("stop btn");
socket.emit("myapp:search", function(data) {
console("stopped stream" + data);
});
});
}
错误节点控制台: TypeError:_stream.abort不是函数
Desktop/assign2/routes/index.js:41
_stream.abort();
错误浏览器控制台:
The connection to ws://localhost:3000/socket.io/?EIO=3&transport=websocket&sid=ZZusoZ2GTYEGTebcAAAB was interrupted while the page was loading.
Source map error: TypeError: NetworkError when attempting to fetch resource.
Resource URL: http://localhost:3000/socket.io/socket.io.js
Source Map URL: socket.io.js.map[Learn More]
答案 0 :(得分:1)
在文件的开头,脚本将_stream设置为空对象。因此,它稍后尝试调用的方法不存在。