CODEWITHSUNDEEP
javascriptrequirejsamdr.js
null_radix
null_radix

Reputation: 702

combine nested modules with require.js optimizer

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.

Upvotes: 1

Views: 1482

Answers (2)

null_radix
null_radix

Reputation: 702

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'
}

Upvotes: 1

kryger
kryger

Reputation: 13181

r.js (I'm assuming that's what you mean by "optimizer") won't touch c and d unless you explicitly ask it to ["You can always explicitly add modules that are not found via the optimizer's static analysis by using the include option." - from RequireJS docs]. If you do that they'll end up in the optimized output and will be loaded at runtime together with a and b and the rest of the code (even if x is false and the require block is never reached).

If you don't, r.js will skip that block and RequireJS will only resolve c and d at runtime (if x is true), when it's too late to do any optimizations or concatenations.

RE: the edit

The correct way to force including c and d in main would be:

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

This means: "include everything main.js requires explicitly but also add c.js and d.js"

RE: the comment:

Oh, you're right, your configuration will also create an optimised c.js that includes both c and d but RequireJS will load the 'standalone' d anyway. Strangely, it will still use the d from the optimised file and won't even evaluate the standalone d.js (checked by editing contents of optimised c.js and d.js). I guess RequireJS requests all listed dependencies immediately since it has no way of knowing that loading c will satifsy the d dependency (this is only known after the module is loaded and processed). That makes sense because the only other option would be request the files sequentially, one by one - that would be a bottleneck.

Upvotes: 2

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