typescript check if type has key

Our problem is classic in Typescript: an object type is undifferentiated, and we would like to differentiate the different cases. No need to check if the prop exists, as the typeof of an undefined prop is 'undefined'. That seems to more reflect what you're trying to do here. Different properties on an object can have totally different types, and we don't even know what obj looks like. Several of my projects include cloud functions that will be accessed by my team or even third parties. runtypes and io-ts are two other alternatives. This article and 44 others are part of the TypeScript Evolution series. Consider the following gist: The type guard is definitively wrong, and yet, Typescript is OK. /u/nadameu is right.

Need to use Object.keys instead Object.values. This Record is how Typescript defines an object with any values for its keys. Recommended: Doing this manually is a ton of work, and if you're dealing with unknown objects often it would be better to lean onto a validation library to simplify it. My personal favorite is superstruct, although lots of people like joi or zod. It represents the type of the property K of the type T. If we now access the three todo properties via the prop method, each one will have the correct type: Now, what happens if we pass a key that doesn't exist on the todo object? // person = { } & Record<'name', unknown>, // do something with person.name, which is a string. Theres no need to explicitly define the generics, theyre getting inferred by usage. With Record this would correctly be a compiler error within the type guard. Im generally either working with data generated by me or the team, or with a third party where Ive tested and verified and know the internal shape of the object, so I can just type cast. A TypeScript enum is compiled to an object at runtime: One benefit of this approach is being able to check at runtime whether a string is a member of the enum: The alternative for enum is using a union type: Type information doesn't exist at runtime, how can we check at runtime if a string is a member of the union type? Instead, we'll require that the key actually exists on the type of the object that is passed in: TypeScript now infers the prop function to have a return type of T[K], a so-called indexed access type or lookup type. Bear in mind Im self-taught and some of this might be outside the norm but Ive found it to be helpful. Here's a first attempt: With these two type annotations in place, obj must be an object and key must be a string. If you leave it off, the compiler will use its default type inference behavior, which will possibly result in a wider or more general type. If the input werent valid it wouldnt have made it this far, so we can truck on without checking it. Why wouldn't typescript allow me to use 'foo' in x and then just expand the type with & { foo: unknown } if it returns true is beyond me. Check out I too wish type inference considered more checks like in, but some pain can be avoided by using more generic type guards instead of writing specialized ones for every case: They are usually written once and can be used throughout the whole project or moved to a library. Heres a playground for you to fiddle around. Are you sure you want to hide this comment? TypeScripts control flow analysis lets you narrow down from a broader type to a more narrow type: This is a type-safety check in JavaScript, and TypeScript benefits from that. Seconding the recommendation to use a library for this. It can be tricky sometimes to capture the semantics of certain operations in a static type system. We've now restricted the set of possible values for both parameters. Your isRecord is incorrect. Also, you could make the return type a little broader: Record. There are plenty of square brackets involved in the return type, admittedly, but there's the type safety we've been looking for! How can we do this automatically? Avid skier This is prone to errors because we might forget to update the array or the type. Here is my current code: What I am trying to do here is that I am using the JavaScript in operator to check that the key name is in the object. When you add or remove members, we need to update the array. TypeScript at the time of this writing needs a little bit more assistance from us. We can now use this variable: A type guard has to be correct, otherwise it can introduce errors. DEV Community 2016 - 2022. Tangent: This is part of why I don't understand why people think these type-predicate-returning functions are so cool. I'll update the blogpost. Which is a lot nicer than using an enum! Lets assume you have a JavaScript object where you dont know if a certain property exists. Validating data as early as possible and disallowing invalid states on the type level is one of the pillars of robust code. It queries the set of keys for a given type, which is why it's also called an index type query. Oh, you're right. It's necessary due to how TypeScript defines the object type, which is basically anything that is not a primitive. However, it gives me the following error at object['name']: I would assume TypeScript would know that the key name is in the object because I used the in operator to check this but it gives me an error. With that definition, it won't allow you to access any properties since it doesn't know about them. Built on Forem the open source software that powers DEV and other inclusive communities. And in our case, both Person and Animal are interfaces, and there is no way instanceof will work at runtime. But at runtime, undefined appear. They all do the same thing, To answer your question: The typeof object === 'object' actually doesn't type guard anything, it is still an object. I don't want to write 5 type guards, casts and inline types to just get one attribute out of a thrown error to print it. Once unpublished, this post will become invisible to the public Thanks for spotting! We define the possible options using an array with a const assertion. Once suspended, hansott will not be able to comment or publish posts until their suspension is removed. With a Record however, it allows you to access string keys. Our newsletter gives you links, updates on fettblog.eu, conference talks, coding soundtracks, and much more.

In so doing, I can eliminate most validation after the initial entry point. Shoot a tweet! Templates let you quickly answer FAQs or store snippets for re-use. Equipped with keyof, we can now improve the type annotations of our prop function. In the example below, we have two interfaces, Animal and Person: But if we want to print cat, dog or human, it becomes more complicated Below, both functions dont work: The first one because type is not defined, the second one because instanceof works with classes. However, there are some cases where Even though this works with JavaScript. The compiler complains, and that's a good thing! and nullish coalescing (??) Press question mark to learn the rest of the keyboard shortcuts. Was this helpful? If you want to know more, lets check out whats happening: Thats it! We no longer want to accept arbitrary strings for the key parameter. Strong interest in domain-driven design and functional programming, Full Stack Functional Programmer focused on Accessibility, Testing, and Documentation/Technical Writing. In those, after authentication I check to the existence of the specific objects and properties required by that function before moving forward, and send back an error if anything is not the right shape. Take a simple prop function, for instance: It accepts an object and a key and returns the value of the corresponding property. Piano & guitar student Got a comment? With you every step of your journey. We're a place where coders share, stay up-to-date and grow their careers. If hansott is not suspended, they can still re-publish their posts from their dashboard. I guess in enum case have a bug. You should use Record instead of Record though. Coming from a JS background, checking the type of an object in Typescript is kind of obscure at first. This is definitely the best practice. We need to provide a little more type information to make that possible. operators, assertion functions, truly private class fields, conditional types, template literal types, adn more. Posted on Jul 12, 2021 Leaving a small tip helps me a lot! We are used to if(obj.property) {//obj.property exists here !} Thank you so much for the explanation! , Let me know if this blogpost was useful! Enter TypeScript 2.1 and the new keyof operator. For further actions, you may consider blocking this person and/or reporting abuse.

A deep dive into the fundamnetals of TypeScripts type system. Typescript allows us to create our own function, that it will understand, and will narrow the type in the following code (in terms of scope, of course). How can I resolve this? He/Him, Simple way to serialize objects to JSON in TypeScript. A const assertion tells the compiler to infer the narrowest or most specific type it can for an expression.

They can still re-publish the post if they are not suspended. typeof obj === "object" narrows to object | null. Leaving a small tip helps me a lot! The object might be any or unknown. For some inane reason, typeof null === 'object' in JavaScript/TypeScript. Need help?

In JavaScript, you would check for properties like that: At the moment, TypeScript isnt able to extend the type of obj with a prop. Have a look! I have an object which I want to validate is the same structure/shape as some interface. 3-4 updates per month, no tracking, spam-free, hand-crafted. Learn about the optional chaining (?.) The return type is still inferred to be any, however: Without further information, TypeScript can't know which value will be passed for the key parameter, so it can't infer a more specific return type for the prop function. A lovely little helper to make TypeScript understand your code better. Updated on May 7, There's some ongoing debate whether enum in TypeScript should be used: The Dangers of TypeScript Enums. TypeScript is a typed superset of JavaScript that compiles to plain JavaScript. TypeScript is a language for application-scale JavaScript development. We can, however, write a little helper function to get correct typings: If you dont want to know how this works, copy it and be happy. Press J to jump to the feed. With any, you might accidentally add something like this to the conditions, which would be OK by the compiler (because of any) but would crash at runtime. Confessions of a Reluctant Ionic-React Fan, Object-Oriented Programming with TypeScriptEncapsulation, function typeGuard(toBeDetermined: any): toBeDetermined is Animal {}, const tg = (tbd: any): tbd is Animal => {//return true if Animal}, http://www.typescriptlang.org/docs/handbook/advanced-types.html. Made with love and Ruby on Rails. And you see the too classic cannot read property 'alpha' of undefined coming, If you want to learn more about typeguards, the official documentation can be a good starting point, although, as of today, it uses the good old Typescript 2.0s typecast <>.

Startup co-founder It will become hidden in your post, but will still be visible via the comment's permalink. Hope it helped ! DEV Community A constructive and inclusive social network for software developers. Once unpublished, all posts by hansott will become hidden and only accessible to themselves. The key here is the type of this function: toBeDetermined is Animal . Let's assume we have defined the following Todo interface: We can apply the keyof operator to the Todo type to get back a type representing all its property keys, which is a union of string literal types: We could've also written out the union type "id" | "text" | "due" manually instead of using keyof, but that would've been cumbersome, error-prone, and a nightmare to maintain. You would need to create a separate one that casts unknown -> Record which then you can start examining properties. A type guard is some expression that performs a runtime check that guarantees the type in some scope. This is such an inconveniet thing to do in typescript I often do a as never cast to just shut up the errors. Trying to be less clever and more kind. Don't know why it was necessary here, but it fixed the error. and only accessible to Hans Ott (journy.io). Was this helpful? Once unsuspended, hansott will be able to comment and publish posts again. You guessed, this is what this article is about :). It prevented us from trying to read a property that's not there. For another real-world example, check out how the Object.entries() method is typed in the lib.es2017.object.d.ts type declaration file that ships with the TypeScript compiler: The entries method returns an array of tuples, each containing a property key and the corresponding value. I've written a book on TypeScript! Typescript understands that if we return true, the argument is an Animal. So how could we type this function in TypeScript? Hi, I am new to TypeScript. JavaScript is a highly dynamic language. I found this very useful! I am using a type guard function to do this. and this is not possible in Typescript, In some cases, you can use type guards. Can you just change the type of the isCity argument from unknown to any?

Maybe not relevant to your application, but I like defensively clearing out incompatible objects before they get into the core of my code.

It's just a type cast, and it's so easy to screw them up that they really should be avoided unless very necessary. Then I am trying to see if that key name is of type string. TypeScript in 50 Lessons, published by Smashing Magazine. . If you want to write secure type guards, for example validating objects fetched from an API, you can check: Now you know how to narrow a type of an object :). (source: http://www.typescriptlang.org/docs/handbook/advanced-types.html). Also, it would've been a solution specific to the Todo type rather than a generic one.