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

Question

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?

Answer 1

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.