javascript - 让 Gulp watch 仅在更改的文件上执行功能

Question

我是 Gulp 的新手并且有以下 Gulpfile

var gulp = require('gulp');
var jshint = require('gulp-jshint');
var concat = require('gulp-concat');
var rename = require('gulp-rename');
var uglify = require('gulp-uglify');

gulp.task('compress', function () {
    return gulp.src('js/*.js') // read all of the files that are in js with a .js extension
      .pipe(uglify()) // run uglify (for minification)
      .pipe(gulp.dest('dist/js')); // write to the dist/js file
});

// default gulp task
gulp.task('default', function () {

    // watch for JS changes
    gulp.watch('js/*.js', function () {
        gulp.run('compress');
    });

});

我想将其配置为重命名、缩小并仅将更改后的文件保存到 dist 文件夹中。做这个的最好方式是什么？

Answer 1

是这样的:

// Watch for file updates
gulp.task('watch', function () {
    livereload.listen();

    // Javascript change + prints log in console
    gulp.watch('js/*.js').on('change', function(file) {
        livereload.changed(file.path);
        gutil.log(gutil.colors.yellow('JS changed' + ' (' + file.path + ')'));
    });

    // SASS/CSS change + prints log in console
    // On SASS change, call and run task 'sass'
    gulp.watch('sass/*.scss', ['sass']).on('change', function(file) {
        livereload.changed(file.path);
        gutil.log(gutil.colors.yellow('CSS changed' + ' (' + file.path + ')'));
    });

});

也很好用 gulp-livereload使用它，您需要安装 Chrome plugin顺便说一句，让它工作。