I am facing a weird issue. In my (lets say) a.ts I have -
/// <reference path="../typings/mocha/mocha.d.ts" />
/// <reference path="../typings/should/should.d.ts" />
import should = require('should');
import something_else = require('../something-else');
Now when I compile using command -
tsc -m commonjs --outDir "./build" "src/test/a.ts"
My generated javascript is not having require for should -
/// <reference path="../typings/mocha/mocha.d.ts" />
/// <reference path="../typings/should/should.d.ts" />
var service_manager = require('../routes/service-manager');
This seems like a bug in typescript compiler, but I may be doing it incorrectly. Or if there is some workaround, please share.
It does that because you are not using it. It will stick as soon as you you actually use the should variable. e.g.
/// <reference path="../typings/mocha/mocha.d.ts" />
/// <reference path="../typings/should/should.d.ts" />
import should = require('should');
var persist = should;
Reason: It allows you to use type information on its own without taking a runtime dependency on require('should'). It also allows you to do lazy loading in AMD scenarios.
As the comment by Eric Nicholson, just require without import.
require('should');
// use should
Besides, those reference path would be bundled at typings/tsd.d.ts by default and no need to write in individual file.