Question

基本上我想看看我的jekyll项目，当发生变化时，我的浏览器会刷新以反映它们。我的想法是将gulp-jekyll集成到我的工作流程中，这样做也可以启动服务器......

var gulp = require('gulp');
gutil = require('gulp-util'),
sass = require('gulp-ruby-sass'),
autoprefixer = require('gulp-autoprefixer'),
minifycss = require('gulp-minify-css'),
uglify = require('gulp-uglify'),
concat = require('gulp-concat'),
notify = require('gulp-notify'),
connect = require('gulp-connect'),
exec = require('gulp-exec'),

livereload = require('gulp-livereload'),
lr = require('tiny-lr'),
server = lr();

var jsSources = ['source/assets/js/*.js', 'source/assets/lib/*.js'];
var sassSources = ['source/assets/sass/style.scss'];

    gulp.task('styles', function() {
    return gulp.src(sassSources)
        .pipe(sass({
            style: 'expanded'
    }))
        .pipe(autoprefixer('last 29 version'))
        .pipe(minifycss())
        .pipe(gulp.dest('source/stylesheets/vendor/min'))
        .pipe(connect.reload())
        .pipe(livereload())
        .pipe(notify({
         message: 'By your command..."styles" command executed'
        }))
    });


    gulp.task('js', function() { 
     return gulp.src(jsSources)
         .pipe(uglify()) 
         .pipe(concat('script.js'))
         .pipe(gulp.dest('source/javascript')) 
         .pipe(connect.reload())
         .pipe(livereload())
         .pipe(notify({
          message: 'By your command..."js" command executed'
     }))  
    });

然而，这就是造成错误的原因......请参阅下面的终端输出。

   gulp.task('server', function() {
      connect(connect.static('_site')).listen(9000);
      console.log('Server running at http://localhost:9000/');
      gulp.watch('source/**', function(event) {
      console.log(event.path + ' ' + event.type + ', rebuilding...')
      gulp.src('').pipe(exec("jekyll build && compass compile")).pipe(livereload());
      });
   });

   gulp.task('watch', function() {
     var server = livereload();
     gulp.watch(jsSources, ['js']); 
     gulp.watch(['source/**'], ['server']);
     gulp.watch(['javascript/script.js'], function(e) {
     server.changed(e.path);
   });

    gulp.watch(sassSources, ['styles']);
    });

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

这是终端中的错误......

   [gulp] Starting 'server'...
   [gulp] 'server' errored after 83 μs Object #<Object> has no method 'static'

声明我知道有一个伟大的yeoman generator已经存在。它似乎做了我想要的一切。为什么折磨自己呢？好吧，我实际上自己配置了大部分现有的gulp文件。并且很想弄明白......

尝试将gulp-connect与Jekyll集成

0 个答案: