I'm using two separate gulp tasks to minify templates into js file (1) and concat all js file of the project to the only minified (2).
gulp.task('templates', function() { // RESULT of this task
return gulp.src('templates/**/*.html')
.pipe(dotify({
root : 'templates'
}))
.pipe(concat('templates.js'))
.pipe(gulp.dest('js'));
});
gulp.task('minifyJs', ['templates'], function() {
return gulp.src([
'js/templates.js',
'js/plugins/*.js',
'js/*.js'
])
.pipe(concat('scripts-all.js'))
});
The question is: am I able to avoid creating the templates.js file by processing the result from first task to the second one to concat it with the rest of js's?
Solution: addSrc should be used
return gulp.src('templates/**/*.html')
.pipe(dotify({
root : 'templates'
}))
.pipe(addSrc([
'js/plugins/*.js',
'js/common/*.js',
'js/ui/*.js',
'js/pages/*.js',
'js/*.js'
]))
.pipe(concat('scripts-all.js'))
.pipe(gulp.dest('js/'));