RequireJS multiple module in single file

Question

I want to merge multiple modules into a single file, but I can't find an official document about how to do it. Now I'm using the below method, it works, but I'm wondering about the following:

• Is it technically correct?
• How does RequireJS check whether a module is loaded or not? By using the module name or the file name?
• Will this solution init modules multiple times in some situation?

index.html

<script src="require.js"></script>
<script>
requirejs.config({
    paths: {
        'a': 'test',
        'b': 'test'
    }
});

require(['a', 'b'], function () {
    console.log('a & b loaded');
});
</script>

test.js

console.log('loading test.js');
// I have some init here, like "Avoid `console` errors in IE"

define('a', function () {
    console.log('a module');
});

define('b', function () {
    console.log('b module');
});

Answer 1

Shangan is right. In http://requirejs.org/docs/api.html#define there is a part that says:

There should only be one module definition per file on disk.

But if you want to do it anyway I found something like that: It is from official documentation: http://requirejs.org/docs/api.html#config-bundles

It says that you can define in which file you can find which modules by bundles property. For example: I have file called modules.js with such content:

define('module_one', function(){
    return 'Hello. I am module one';
})

define('module_two', function(){
    return 'Hello. I am module two';
})

Then in my main.js file I do something like this:

requirejs.config({
    baseUrl : 'modules/',
    bundles: {
        'modules': ['module_one', 'module_two']
    }
})

require(['module_one', 'module_two'], function(moduleOne, moduleTwo){
    alert(moduleOne);
    alert(moduleTwo);
})

And it works like I want to.