如何包含适当的JavaScript文件的版本（精简版还是原始版）？

Question

I am developing a webapp (if it matters, I use npm and bower ). My HTML file includes angular.js . Now, I want that my production-version of the webapp uses the minified version angular.min.js .

Considering I do want two versions of my webapp:

• A "raw" version which includes angular.js
• A "production" version which includes angular.min.js

How is it traditionally done given that I want to maintain a single version of index.html ?

Answer 1

You can either do it manually or you can use Gulp or Grunt

They will help you to get your files ready for the production environment.

This tutorial was a big help when I started with gulp.

Answer 2

I opted for sed invoked in a Makefile :

$(RELEASE_DIR)/index.html: $(PUBLIC_DIR)/index.html | $(RELEASE_DIR)/
    $(SED) 's/\.js/.min.js/g' $< > $@

$(RELEASE_DIR)/:
    $(MKDIR) --parents $@

I am accepting my own answer because it was simpler to use sed rather than using yet another dependency (I already had a Makefile ).