Question

我有来自https://github.com/gulpjs/gulp的gulpfile.js。我只是改变我的文件夹名称。任务'gulp'正常工作，首先'gulp watch'。但任何其他保存任何较少的文件都不会做任何事情。仅仅在1.34秒后“完成'风格'” （JS工作得很好）

有什么建议吗？ main.less与所有filles的进口量减少文件夹看起来像：

/www
    /js
    /css
        main.less
        file.less
        file2.less
        /lib
            file3.less




var gulp = require('gulp');
var less = require('gulp-less');
var babel = require('gulp-babel');
var concat = require('gulp-concat');
var uglify = require('gulp-uglify');
var rename = require('gulp-rename');
var cleanCSS = require('gulp-clean-css');
var del = require('del');

var paths = {
    styles: {
        src: 'www/css/main.less',
        dest: 'www/css/'
    },
    scripts: {
        src: 'www/js/js/**/*.js',
        dest: 'www/js/'
    }
};

function clean() {
  return del([ 'assets' ]);
}

function styles() {
  return gulp.src(paths.styles.src)
    .pipe(less())
    .pipe(cleanCSS())
    // pass in options to the stream
    .pipe(rename({
      basename: 'main',
      suffix: '.min'
    }))
    .pipe(gulp.dest(paths.styles.dest));
}

function scripts() {
  return gulp.src(paths.scripts.src, { sourcemaps: true})
    .pipe(babel())
    .pipe(uglify())
    .pipe(concat('all.min.js'))
    .pipe(gulp.dest(paths.scripts.dest));
}

function watch() {
  gulp.watch("www/js/js**/*.js", scripts);
  gulp.watch("www/css/**/*.less", styles);
}

exports.clean = clean;
exports.styles = styles;
exports.scripts = scripts;
exports.watch = watch;

var build = gulp.series(clean, gulp.parallel(styles, scripts));

gulp.task('build', build);

gulp.task('default', build);

Answer 1

正如@Jakub所想，你实际上从未在你提供的代码中调用你的手表任务！你可能想要这样的东西：

var build = gulp.series(clean, gulp.parallel(styles, scripts), watch);

Gulp手表 - 不工作

1 个答案: