Question

I have the following.

require([ 'a', 'b' ], function() {     
   if(x){
       require([ 'c', 'd' ], function( ) {});
   }   
});

c will always be loaded with d. Is there anyway to make the optimizer optimize c and d so that they will load as one file?

EDIT:

Also im using r.js as the optimizer. And the modules for my build script look something like this.

modules: [
    {
        name: "main",
    },
    {
        name: "c",
            include: ["d"]
    }
] 

This will include d in c, but both c and d are still loaded at runtime.

Answer 1

I got the answer here

You can set up a path mapping in production to inform RequireJS that d is located at c.

Simply do:

paths: {
  'd': 'c'
}