Question

我正在尝试使用gulp

使用快速服务器进行实时重新加载

然而gulp在运行后退出如下，并且无法点击服务器

cmd输出：

D：\ Users \ workspace \ server＆gt; gulp服务器   [11:31:56]使用gulpfile D：\ Users \ workspace \ server \ gulpfile.js

[11:31:56]开始＆＃39;服务器＆＃39; ...

[11:31:56]已完成的服务器＆＃39; 26毫秒后

livereload [tiny-lr]收听35729 ...

d：\ Users \用户工作空间\服务器＆GT;

gulpfile.js：

var gulp = require('gulp');
var server = require('gulp-express');

gulp.task('server', function () {
   // Start the server at the beginning of the task 
server.run(['app.js']);

// Restart the server when file changes 
gulp.watch(['app/**/*.html'], server.notify);
gulp.watch(['app/styles/**/*.scss'], ['styles:scss']);
//gulp.watch(['{.tmp,app}/styles/**/*.css'], ['styles:css', server.notify]); 
//Event object won't pass down to gulp.watch's callback if there's more than one of them. 
//So the correct way to use server.notify is as following: 
gulp.watch(['{.tmp,app}/styles/**/*.css'], function(event){
    gulp.run('styles:css');
    server.notify(event);
    //pipe support is added for server.notify since v0.1.5, 
    //see https://github.com/gimm/gulp-express#servernotifyevent 
});

gulp.watch(['app/scripts/**/*.js'], ['jshint']);
gulp.watch(['app/images/**/*'], server.notify);
gulp.watch(['app.js', 'routes/**/*.js'], [server.run]);
});

gulp.task('default', ['server']);

gulpjs tutplus

我称之为gulp的方式是否存在根本不正确的事情？

Answer 1

即使我有一些问题让实时重新加载工作。因此我开始使用gulp-nodemon来创建网络服务器和实时重新加载。以下是我的任务：

    gulp.task('server', function(){  
        var nodeOptions = {
            script:"",//path to the app.js or server.js 
            delayTime: 1,
            env: {
                'PORT': 8088 //port that you want to run it on
            },
            watch: ['filesToBeWatched.js'] //files to be watched for changes
        };

        return nodemon(nodeOptions)
            .on('restart', function(event){
                console.log('Node Server RESTARTED');
            })
            .on('start', function(){
                console.log('Node Server STARTED');
                startBrowserSync();
            })
            .on('crash', function() {
                console.log('Node Server CRASHED');
            })
            .on('exit', function() {
                console.log('Node Server EXISTED')
            });

    });

Gulp在跑完后退出

1 个答案: