Gulp手表在文件更改时未触发缩小

Question

我有一个简单的入门应用程序，我创建了gulpfile.js文件，其内容如下，

let gulp = require('gulp');
let cleanCSS = require('gulp-clean-css');
// Task to minify css using package cleanCSs
gulp.task('minify-css', () => {
     // Folder with files to minify
     return gulp.src('src/assets/styles/*.css')
     //The method pipe() allow you to chain multiple tasks together 
     //I execute the task to minify the files
    .pipe(cleanCSS())
    //I define the destination of the minified files with the method dest
    .pipe(gulp.dest('src/assets/dist'));
});

//We create a 'default' task that will run when we run `gulp` in the project
gulp.task('default', function() {
// We use `gulp.watch` for Gulp to expect changes in the files to run again
  gulp.watch('./src/assets/styles/*.css', function(evt) {
    gulp.task('minify-css');
  });
});

如果运行gulp minify-css，它可以正常工作，但是我需要它来减小文件更改的大小

但是所有的操作都会在cmd窗口中记录一条消息，例如'Starting ...'

我什至不知道这是什么意思...

package.json：

..
     "gulp": "^4.0.2",
     "gulp-clean-css": "^4.2.0"

Answer 1

我认为您在运行任务minify-css时需要添加return，以便系统知道上一个任务何时完成。

gulp.task('default', function() {
// We use `gulp.watch` for Gulp to expect changes in the files to run again
  gulp.watch('./src/assets/styles/*.css', function(evt) {
    return gulp.task('minify-css');
  });
});