如何将整个文件夹及其子文件夹gulp.src，然后基于文件gulp.dest'；地点？

编辑：是否有方法清理此代码

咖啡

# Watch pages
gulp.task 'jade', ->
  # Watch index
  gulp.src('src/jade/index.jade')
  .pipe(jade(pretty: true))
  .pipe gulp.dest('dist')
  # Watch views
  gulp.src('src/jade/views/*.jade')
  .pipe(jade(pretty: true))
  .pipe gulp.dest('dist/views')
  # Watch views/products
  gulp.src('src/jade/views/products/*.jade')
  .pipe(jade(pretty: true))
  .pipe gulp.dest('dist/views/products')

gulp.watch 'src/jade/*.jade', ['html']
gulp.task 'html', (callback) ->
  runSequence 'jade', callback
  return

# Watch pages
gulp.task 'jade', ->
  gulp.src('src/jade/*.jade')
  .pipe(jade(pretty: true))
  .pipe gulp.dest('dist')
gulp.watch 'src/jade/*.jade', ['html']
gulp.task 'html', (callback) ->
  runSequence 'jade', callback
  return

---

假设我正在运行gulp任务来处理我的.jade文件，并且我正在开发一个angular应用程序（views/***/.html），我如何保持任务干净，以便将任务更改为这样做

// gulp.src('src/jade/**/*.jade')
// gulp.dest('dist/path/*.html') so for example 'src/jade/index.jade'
// will be output into 'dist/index.html' and
// 'src/jade/views/products/product.jade' will be
// output into 'dist/views/products/product.html'

咖啡

# Watch pages
gulp.task 'jade', ->
  # Watch index
  gulp.src('src/jade/index.jade')
  .pipe(jade(pretty: true))
  .pipe gulp.dest('dist')
  # Watch views
  gulp.src('src/jade/views/*.jade')
  .pipe(jade(pretty: true))
  .pipe gulp.dest('dist/views')
  # Watch views/products
  gulp.src('src/jade/views/products/*.jade')
  .pipe(jade(pretty: true))
  .pipe gulp.dest('dist/views/products')

gulp.watch 'src/jade/*.jade', ['html']
gulp.task 'html', (callback) ->
  runSequence 'jade', callback
  return

# Watch pages
gulp.task 'jade', ->
  gulp.src('src/jade/*.jade')
  .pipe(jade(pretty: true))
  .pipe gulp.dest('dist')
gulp.watch 'src/jade/*.jade', ['html']
gulp.task 'html', (callback) ->
  runSequence 'jade', callback
  return

task.js

gulp.task('jade', function() {
  return gulp.src('src/jade/*.jade').pipe(jade({
    pretty: true
  })).pipe(gulp.dest('dist'));
});

gulp.watch('src/jade/*.jade', ['html']);

gulp.task('html', function(callback) {
  runSequence('jade', callback);
});

---

您的问题的答案已经在您自己的帖子中：

// gulp.src('src/jade/**/*.jade')

在您的

jade

任务和

watch

中使用此选项应该可以完全完成您想要的：

gulp.task('jade', function() {
  return gulp.src('src/jade/**/*.jade')
    .pipe(jade({pretty: true}))
    .pipe(gulp.dest('dist'));
});

gulp.watch('src/jade/**/*.jade', ['html']);

这将在

dist

文件夹中生成如下文件：

src/jade/index.jade -> dist/index.html
src/jade/views/example.jade -> dist/views/example.html
src/jade/views/products/product.jade -> dist/views/products/product.html
...

---

每个文件是否会自动输出到相应的路径中？gulp.dest（'dist'）是否只是将'src/path/of/the/*.file'更改为'dist/path/of/the/*.file'？是的，请参阅我编辑的答案。

**

之前的所有内容都会被剥离（即

src/jade/

），但

**

之后的文件夹结构会保留下来。