如何将未缩小和缩小的js文件与browserfy捆绑在一起

Question

我得到了一个缩小的捆绑包文件，我想在同一个dist中得到一个非缩小的捆绑。

gullfile.js

'use strict';

var gulp        = require('gulp'),
    gulpLoad    = require('gulp-load-plugins'),
    browserify  = require('browserify'),
    source      = require('vinyl-source-stream'),
    buffer      = require('vinyl-buffer'),
    del         = require('del'),
    pkg         = require('./package.json'),
    $           = gulpLoad(),
    DIST        = './dist',
    SRC         = './src/index.js';

gulp.task('clean', function(fn) {
  return del(DIST, fn);
})

gulp.task('lint', function() {
  return gulp.src(SRC)
    .pipe($.jshint())
    .pipe($.jshint.reporter('default'))
});

gulp.task('bundle', ['lint', 'clean'], function() {
  var b = browserify();
  return b.bundle()
    .pipe(source('./ar-string.min.js'))
    .pipe(buffer())
    .pipe($.sourcemaps.init({loadMaps: true}))
        .pipe($.uglify())
        .on('error', $.util.log)
    .pipe($.sourcemaps.write('./'))
    .pipe(gulp.dest(DIST));
});

gulp.task('default', function() {
  gulp.watch([SRC, './gulpfile.js'], ['bundle']);
});

Answer 1

EDIT：事实上，我不认为这可以在一个管道中完成，我们可以有两个，一个不缩小，一个缩小和源映射：

b.bundle()
.pipe(source('./ar-string.js'))
.pipe(gulp.dest(DIST));
return b.bundle()
.pipe(source('./ar-string.min.js'))
.pipe(buffer())
.pipe($.sourcemaps.init({loadMaps: true}))
    .pipe($.uglify())
    .on('error', $.util.log)
.pipe($.sourcemaps.write('./'))
.pipe(gulp.dest(DIST));