node.js - Express js 4 上的 GET/easyrtc/easyrtc.js 404

Question

我在 Express.js 4 上遇到 EasyRTC 404 问题。控制台显示:GET/easyrtc/easyrtc.js 404 4.500 ms - 1077 这是我的 app.js 代码:

var express = require('express');
var path = require('path');
var favicon = require('serve-favicon');
var logger = require('morgan');
var cookieParser = require('cookie-parser');
var bodyParser = require('body-parser');
var io = require('socket.io');
var easyrtc = require('easyrtc');

var routes = require('./routes/index');
var users = require('./routes/users');

var staticMiddleware = express.static(__dirname + "/public");

var app = express();

// view engine setup
app.set('views', path.join(__dirname, 'views'));
app.set('view engine', 'ejs');

// uncomment after placing your favicon in /public
//app.use(favicon(__dirname + '/public/favicon.ico'));
app.use(logger('dev'));
app.use(bodyParser.json());
app.use(bodyParser.urlencoded({ extended: false }));
app.use(cookieParser());
app.use(express.static(path.join(__dirname, 'public')));

app.use('/', routes);
app.use('/users', users);
app.get('/easyrtc/*', function (req,res) {
    res.sendFile(__dirname + '/static');
})
// catch 404 and forward to error handler
app.use(function(req, res, next) {
    var err = new Error('Not Found');
    err.status = 404;
    next(err);
});

// error handlers

// development error handler
// will print stacktrace
if (app.get('env') === 'development') {
    app.use(function(err, req, res, next) {
        res.status(err.status || 500);
        res.render('error', {
            message: err.message,
            error: err
        });
    });
}

// production error handler
// no stacktraces leaked to user
app.use(function(err, req, res, next) {
    res.status(err.status || 500);
    res.render('error', {
        message: err.message,
        error: {}
    });
});


app.set('port', process.env.PORT || 3000);

var server = app.listen(app.get('port'), function() {
  console.log('Express server listening on port ' + server.address().port);
});

var webSockets = io.listen(server);
var rtc = easyrtc.listen(app, webSockets);

请帮助我，我一直在寻找解决方案，但我根本找不到解决方案。先谢谢了

注意:我正在从 Linux 中的终端生成express.js 项目，使用命令:express -e

Answer 1

好吧，您可以删除下面的代码:

app.use(function(req, res, next) {
  var err = new Error('Not Found');
  err.status = 404;
  next(err);
});