javascript - 必须多次运行 gulp 才能获得样式更改

Question

我在编译我的 CSS 时遇到问题，当我在我的 SASS 文件中进行更改时，它不会更改最终文件，只是在尝试了不止一次之后。

    var gulp = require('gulp');
    var sass = require('gulp-sass');
    var rename = require('gulp-rename');
    var postcss = require('gulp-postcss');
    var autoprefixer = require('autoprefixer');
    var cssnano = require('cssnano');
    var concat = require('gulp-concat');
    var uglify = require('gulp-uglify');

    //CSS Tasks
    gulp.task('styles', function() {
        console.log("Compilling SASS");
        gulp.src('app/sass/style.scss')
        .pipe(sass().on('error', sass.logError))
        .pipe(postcss([ autoprefixer() ]))
        .pipe(rename('999_style.css'))
        .pipe(gulp.dest('app/css/'));
    });
    gulp.task('stylescompress', ['styles'], function() {
        console.log("Concating and moving all the css files in styles folder");
        gulp.src("app/css/**.css")
        .pipe(concat('style.css'))
        .pipe(postcss([ cssnano() ]))
        .pipe(gulp.dest('css/'));
    });

    //Javascript Tasks
    gulp.task('jscompress', function() {
        //console.log("Concating and moving all the js files in javascript folder");
        gulp.src("app/js/**.js")
        .pipe(concat('scripts.js'))
        .pipe(gulp.dest('js/'))
        .pipe(rename('scripts.min.js'))
        .pipe(uglify())
        .pipe(gulp.dest('js/'))
    });

    //Watch task
    gulp.task('default',function() {
        gulp.watch(['app/sass/**/*.scss'],['stylescompress'])
        .on('change', function(event) {
            console.log('SASS - File ' + event.path + ' was ' + event.type + ', running tasks...');
        });
        gulp.watch(['app/js/**/*.js'],['jscompress'])
        .on('change', function(event) {
            console.log('SCRIPTS - File ' + event.path + ' was ' + event.type + ', running tasks...');
        });
    });

如你所见，我根据Gulp Docs使用了all，但我并没有发现这是怎么回事。

Answer 1

我错过了它完成时的返回，让 Gulp 了解它应该调用下一个任务的时间，它记录在 Gulp Docs 中

我使用流让 gulp 理解每个任务何时结束，如下所示:

gulp.task('somename', function() {
  var stream = gulp.src('client/**/*.js')
    .pipe(minify())
    .pipe(gulp.dest('build'));
  return stream;
});