编辑:有没有办法清除此代码?
任务咖啡
# 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
假设我正在运行gulp任务来处理.jade文件,而我正在开发一个有角度的应用程序(views / ** / *。html),如何保持任务清洁以更改任务以执行这个?
// 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'
任务咖啡
# 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
任务中使用它,watch
应该完全完成您想要的操作:
gulp.task('jade', function() {
return gulp.src('src/jade/**/*.jade')
.pipe(jade({pretty: true}))
.pipe(gulp.dest('dist'));
});
gulp.watch('src/jade/**/*.jade', ['html']);
这样将在dist
文件夹中生成文件,如下所示:
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
...
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