逐个运行gulp命令会产生与在gulp.series中运行它们不同的输出。为什么？

Question

下面是我的gulpfile。我尝试从scss生成简化的css，然后使用gulp-rev.重命名。运行gulp clean:assets，然后gulp scss，然后gulp css完美工作。但是运行gulp build并不能完成这个任务。缩小的css将存储在./ gets / css中，但重命名的css不会存储在./public/gets中

const gulp = require('gulp');
//gulp sass converts sass to css
const sass = require('gulp-sass');
//cssnano converts css to minified form
const cssnano = require('gulp-cssnano');
//rev renames the assets by appending content hash to filenames so that the browser treats it as a new asset if content is changed
const rev = require('gulp-rev');
const path = require('path');
const del = require('del');

gulp.task('clean:assets', function(done){
    del.sync('./public/assets');
    del.sync('./assets/css');
    done();
});

gulp.task('scss', function(done){
    gulp.src('./assets/scss/**/*.scss')
    .pipe(sass())
    .pipe(cssnano())
    .pipe(gulp.dest('./assets/css'))
    done();
});

gulp.task('css', function(done){
    gulp.src('./assets/css/**/*.css', {base: path.join(process.cwd(), 'assets')})
    .pipe(rev())
    .pipe(gulp.dest('./public/assets/'))
    .pipe(rev.manifest('public/assets/rev-manifest.json',{
        base: process.cwd()+'./public/assets',
        merge: true
    }))
    .pipe(gulp.dest('./public/assets'));
    done();
});

gulp.task('build', gulp.series('clean:assets', 'scss', 'css'), function(done){
    done(); 
})

Answer 1

gulp.task('scss', ()=> {
    return new Promise((resolve, reject) => {
        return gulp.src('./assets/scss/**/*.scss')
            .pipe(sass())
            .pipe(cssnano())
            .pipe(gulp.dest('./assets/css'))
            .on('end', resolve)
            .on('error', reject);
    });
});

将scss任务更改为上面的，以使其正常工作。请参阅https://stackoverflow.com/a/53937286/7374508