Preventing Compilation Errors in Typescript Due to Undeclared Objects

Suppose I have an external JavaScript file included on the page that exports certain properties to the global scope (binds them to window).

I am looking to access some of those properties in my TypeScript projects like this:

UndeclaredExportedProperty.aFunction()

However, TypeScript does not allow me to compile that ><

I don't want to go through the complicated process of declaring the entire interface of the module since I do not know it and frankly, I do not care about it. It is simply a module that I need to call once and "trust" that by the time I make the call, it is loaded and contains the correct elements (it is non-critical, so not calling it will not lead to any major issues).

What is the simplest way to achieve this with TypeScript?

Edit in response to marking as duplicate:

While the answer to that question did resolve my issue, the questions are different (for instance, Stack cannot identify it as a duplicate suggestion) and I believe Pokus' answer provides a more straightforward and general solution for what I am trying to accomplish compared to the answers in that question.

That being said, if an admin deems this still a duplicate or too simplistic of a question, feel free to delete/close it as I have received my answer. Personally, I will keep it up because the next person searching on Google or elsewhere might find a solution more easily this way.

javascripttypescriptwebpackecmascript-6ecmascript-2016

Answer №1

When dealing with external JavaScript libraries, it is recommended to utilize the declare keyword like so:

declare var UndeclaredExportedProperty: any;

This information can be found in the official documentation here: https://www.typescriptlang.org/docs/handbook/modules.html under the section Working with Other JavaScript Libraries

Answer №2

Experiment with

(window as any).UndeclaredExportedProperty.aFunction()

This method will instruct the compiler to consider the Window instance as untyped for a temporary period, granting you flexibility to perform various actions.

Answer №3

Alright, it looks like the answer to this question has been provided in this post:

How can I explicitly add a new property to `window` using TypeScript?

Since I can't mark my own response as a duplicate, I'll just leave this here and close the discussion. Hopefully, anyone searching for the same information in the future will stumble upon this.

Evidently, you can access properties of an object (like window) by using the array selection operator (o.o?). This means you can simply write:

window["UndeclaredExportedProperty"].aFunction()

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