Combine two gulp tasks into one JS file

Question

I have the following two tasks:

gulp.task('compress', () => {
    return gulp.src('app/static/angular/**/*.js')
        .pipe(concat('build.js'))
        .pipe(gulp.dest('./app/static'));
});

gulp.task('templates', () => {
    return gulp.src('app/static/angular/**/*.html')
        .pipe(htmlmin())
        .pipe(angularTemplateCache('templates.js', {
            module: 'myApp',
            root: '/static/angular'
        }))
        .pipe(gulp.dest('./app/static'))
});

And it works fine, but I want them both concatenated into build.js -- how can I combine these two?

Wells · Accepted Answer · 2018-10-10 16:46:30Z

1

In the end I used merge-stream to merge the two streams into one output file:

var gulp = require('gulp');
var concat = require('gulp-concat');
var htmlmin = require('gulp-htmlmin');
var angularTemplateCache = require('gulp-angular-templatecache');
var merge = require('merge-stream');

gulp.task('build', () => {
    var code = gulp.src('app/static/angular/**/*.js');

    var templates = gulp.src('app/static/angular/**/*.html')
        .pipe(htmlmin())
        .pipe(angularTemplateCache({
            module: 'myApp',
            root: '/static/angular'
        }));

    return merge(code, templates)
        .pipe(concat('build.js'))
        .pipe(gulp.dest('./app/static'));
});

gulp.task('default', ['build']);

answered Oct 10, 2018 at 16:46

Wells

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Comments

Rajat · Accepted Answer · 2018-10-10 02:21:07Z

0

I assume the above task mentioned is in separate file say compress.js inside tasks folder

In gulpfile.js you can use below code :

//Include require-dir to include files available in tasks directory

var requireDir = require('require-dir');

// And Take the tasks directory

requireDir('./tasks');

Then you can create a build task as below in gulpfile.js:

gulp.task('build', ['compress', 'templates']);

edited Oct 10, 2018 at 2:21

answered Oct 10, 2018 at 1:59

Rajat

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