小编Yas*_*sin的帖子

我正在学习前端构建系统目前gulp,我想使用brower-sync,问题是它不会在commad行中引发错误,而是当它调出浏览器时它不会显示我的html文件而它会在浏览器窗口中说"无法获取/"错误.这是我的gulpfile.js代码

var gulp = require('gulp'),
   uglify = require('gulp-uglify'),
   compass= require('gulp-compass'),
   plumber = require('gulp-plumber'),
   autoprefixer = require('gulp-autoprefixer'),
   browserSync = require('browser-sync'),
   reload = browserSync.reload,
   rename = require('gulp-rename');


gulp.task('scripts', function() {
   gulp.src(['public/src/js/**/*.js', '!public/src/js/**/*.min.js'])
      .pipe(plumber())
      .pipe(rename({suffix: '.min'}))
      .pipe(uglify())
      .pipe(gulp.dest('public/src/js/'));
});

gulp.task('styles', function() {
   gulp.src('public/src/scss/main.scss')
      .pipe(plumber())
      .pipe(compass({
          config_file: './config.rb',
          css: './public/src/css/',
          sass: './public/src/scss/'
      }))
     .pipe(autoprefixer('last 2 versions'))
     .pipe(gulp.dest('public/src/css/'))
     .pipe(reload({stream:true}));
});


gulp.task('html', function() {
  gulp.src('public/**/*.html');
});  

gulp.task('browser-sync', function() {
    browserSync({
      server: {
         baseDir: "./public/"
      }
   });
});

gulp.task('watch', function() {
   gulp.watch('public/src/js/**/*.js', ['scripts']);
   gulp.watch('public/src/scss/**/*.scss', ['styles']);
   gulp.watch('public/**/*.html', ['html']); …