module.exports in typescript
This has now been implemented and is ready in TypeScript 0.9 :)
So I think I've found a workaround. Just wrap the keyword 'module' in parentheses in your .ts file:
declare var module: any;
(module).exports = MyClass;
The generated javascript file will be exactly the same:
(module).exports = MyClass;
Note, better than declaring var module yourself, download the node.d.ts definition file and stick it in the same directory as your typescript file. Here is a complete sample of an express node.js routing file which assumes node.d.ts is in same directory:
/// <reference path="node.d.ts" />
var SheetController = function () {
this.view = function (req, res) {
res.render('view-sheet');
};
};
(module).exports = SheetController;
I can then new up a SheetController and (using express) assign the view method:
var sheetController = new SheetController();
app.get('/sheet/view', sheetController.view);
I suppose any keyword can be escaped using this pattern:
declare var reservedkeyword: any;
(reservedkeyword).anything = something;
You can export a single class in TypeScript like this:
class Person {
private firstName: string;
private lastName: string;
constructor(firstName: string, lastName: string) {
this.firstName = firstName;
this.lastName = lastName;
}
public getFullName() {
return `${this.firstName} ${this.lastName}`;
}
}
export = Person;
And here is how it's going to be used:
var Person = require('./dist/commonjs/Person.js');
var homer = new Person('Homer', 'Simpson');
var name = homer.getFullName();
console.log(name); // Homer Simpson
To be complete, here is my tsconfig.json (I am using TypeScript v2.0.3):
{
"compilerOptions": {
"module": "commonjs",
"moduleResolution": "node",
"outDir": "dist/commonjs",
"rootDir": "src/ts",
"target": "es5"
},
"exclude": [
"dist",
"node_modules"
]
}