Typescript 1.7+ How to add String prototype function?

Question

Anybody knows how to extend a base class in Typescript 1.7. Something like this example in JS:

String.prototype.foo = function() {
    return 'Bar';
}

I found many ways to do this in older versions of Typescript, like:

interface String {
    foo(): string;
}

String.prototype.foo= function() {
    return 'Bar';
}

or

interface StringConstructor {
    foo(): string;
}

String.foo = function() {
    return 'Bar';
}

I tried many ways to do this, but I aways get error while TS is compiled!

Thank you

Felipe

Are you using external modules? What's the error you get? If you are using external modules then you need to move the interface into a definition file. — David Sherret
– David Sherret, Commented Dec 8, 2015 at 17:01

basarat · Accepted Answer · 2015-12-08 22:57:57Z

1

The following still works in action:

interface String {
    foo(): string;
}

String.prototype.foo= function() {
    return 'Bar';
}

You probably have it in an external module which disconnects it from the global context. More on this : https://basarat.gitbooks.io/typescript/content/docs/types/lib.d.ts.html

answered Dec 8, 2015 at 22:57

basarat

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2 Comments

The unique trick is interface definition should be in a definition file.

It works but produces the following error: TS2339: Property 'foo' does not exist on type 'String' in typescript 1.8