Question

176

How to save a stream into multiple destinations with Gulp.js?

rated 0 times [ 177] [ 1] / answers: 1 / hits: 44171 / 11 Years ago, sat, february 22, 2014, 12:00:00

const gulp = require('gulp');

const $ = require('gulp-load-plugins')();

const source = require('vinyl-source-stream');

const browserify = require('browserify');



gulp.task('build', () =>

  browserify('./src/app.js').bundle()

    .pipe(source('app.js'))

    .pipe(gulp.dest('./build'))       // OK. app.js is saved.

    .pipe($.rename('app.min.js'))

    .pipe($.streamify($.uglify())

    .pipe(gulp.dest('./build'))       // Fail. app.min.js is not saved.

);

Piping to multiple destinations when file.contents is a stream is not currently supported. What is a workaround for this problem?

Answers

Only authorized users can answer the question. Please sign in first, or register a free account.

jaidyn

Add To Favorites

Follow

Total Points: 633

Total Questions: 102

Total Answers: 100

Location: Trinidad and Tobago

Member since Thu, Dec 1, 2022

2 Years ago

answered 11 Years ago janjadonb · Accepted Answer

Currently you have to use two streams for each dest when using file.contents as a stream. This will probably be fixed in the future.

var gulp       = require('gulp');

var rename     = require('gulp-rename');

var streamify  = require('gulp-streamify');

var uglify     = require('gulp-uglify');

var source     = require('vinyl-source-stream');

var browserify = require('browserify');

var es         = require('event-stream');



gulp.task('scripts', function () {

    var normal = browserify('./src/index.js').bundle()

        .pipe(source('bundle.js'))

        .pipe(gulp.dest('./dist'));



    var min = browserify('./src/index.js').bundle()

        .pipe(rename('bundle.min.js'))

        .pipe(streamify(uglify())

        .pipe(gulp.dest('./dist'));



    return es.concat(normal, min);

});

EDIT: This bug is now fixed in gulp. The code in your original post should work fine.