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?
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.