我曾试图在 Gulp 中丑化 Browserify 的产出,但没有奏效。
Gulpfile.js
var browserify = require('browserify');
var gulp = require('gulp');
var uglify = require('gulp-uglify');
var source = require('vinyl-source-stream');
gulp.task('browserify', function() {
return browserify('./source/scripts/app.js')
.bundle()
.pipe(source('bundle.js'))
.pipe(uglify()) // ???
.pipe(gulp.dest('./build/scripts'));
});
据我所知,我不能按照下面的步骤进行。我需要在一个管道中保持序列?
gulp.task('browserify', function() {
return browserify('./source/scripts/app.js')
.bundle()
.pipe(source('bundle.js'))
.pipe(uglify()) // ???
.pipe(gulp.dest('./source/scripts'));
});
gulp.task('scripts', function() {
return grunt.src('./source/scripts/budle.js')
.pipe(uglify())
.pipe(gulp.dest('./build/scripts'));
});
gulp.task('default', function(){
gulp.start('browserify', 'scripts');
});