gulp.src([
// Note: Since we are not using useref in the scripts build pipeline,
// you need to explicitly list your scripts here in the right order
// to be correctly concatenated
'./app/scripts/main.js'
// Other scripts
])
We can add scripts to be concatenated and optimized which means that I should references a single main.min.js file in index.html. But then, I cannot use gulp serve for development because the script file in src/scripts is named main.js (not main.min.js) and there are a bunch of other scripts that are not referenced.
My workaround for now is to serve from dist directory with some watches added to the serve:dist gulp task. I am sure this is not the way WSK meant to be used. So please enlighten me :)
In the
gulpfile
>script
task:We can add scripts to be concatenated and optimized which means that I should references a single
main.min.js
file inindex.html
. But then, I cannot usegulp serve
for development because the script file insrc/scripts
is namedmain.js
(notmain.min.js
) and there are a bunch of other scripts that are not referenced.My workaround for now is to serve from
dist
directory with some watches added to theserve:dist
gulp task. I am sure this is not the way WSK meant to be used. So please enlighten me :)