Typescript check if property in object in typesafe way
Solution 1
You don't get an error because you use a string to check if the property exists.
You will get the error this way:
interface Obj{
a: any;
}
const obj: Obj = { a: "test" };
if (obj.b) // this is not allowed
if ("b" in obj) // no error because you use string
If you want type checking to work for string properties you could add index signatures using this example
Solution 2
The following handle
function checks hypothetical server response typesafe-way:
/**
* A type guard. Checks if given object x has the key.
*/
const has = <K extends string>(
key: K,
x: object,
): x is { [key in K]: unknown } => (
key in x
);
function handle(response: unknown) {
if (
typeof response !== 'object'
|| response == null
|| !has('items', response)
|| !has('meta', response)
) {
// TODO: Paste a proper error handling here.
throw new Error('Invalid response!');
}
console.log(response.items);
console.log(response.meta);
}
Playground Link. Function has
should probably be kept in a separate utilities module.
Solution 3
You can implement your own wrapper function around hasOwnProperty that does type narrowing.
function hasOwnProperty<T, K extends PropertyKey>(
obj: T,
prop: K
): obj is T & Record<K, unknown> {
return Object.prototype.hasOwnProperty.call(obj, prop);
}
I found this solution here: TypeScript type narrowing not working when looping
cdbeelala89
Updated on September 03, 2021Comments
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cdbeelala89 over 2 years
The code
const obj = {}; if ('a' in obj) console.log(42);
Is not typescript (no error). I see why that could be. Additionally, in TS 2.8.1 "in" serves as type guard.
But nevertheless, is there an way to check if property exists, but error out if the property is not defined in the interface of obj?
interface Obj{ a: any; }
I'm not talking about checking for undefined...
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Sean Duggan about 5 yearsPossible duplicate of How to determine whether an object has a given property in JavaScript
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cdbeelala89 about 6 yearsBut "if (obj.b)" also disallows undefined. There is a difference between property not existing and it being undefined. I JUST want to check if the property exists in a typesafe way.
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Kokodoko about 6 yearsThe type system already defines if the property exists or not, so there is no need to check it yourself. Type guards are used when an object can be of multiple types, such as:
const obj: Thing | OtherThing
. -
cdbeelala89 about 6 yearsNo if i have an interface { a?: any; }, I do not know if the property exists. And if I get an object from a server I don't control (e.g.), I never know what property exists. I just want to discriminate in a typesafe way between actual undefined or property not existing at all!
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Kokodoko about 6 yearsIf you don't know what your JSON data looks like, you can use
console.log(o.hasOwnProperty("a"))
to see if a property exists at all. If it exists but it's undefined, then it will still return true. -
cdbeelala89 about 6 yearsstill is not a typesafe check but your answer is stell the best because only one. and i can always write a function to do a typesafe "in" check myself (with typescript keyof syntax)
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Kokodoko about 6 yearsI should add an example of index signatures to get type safety with
obj[“test”]
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Jose almost 4 yearsVery useful to verify interfaces! basarat.gitbook.io/typescript/type-system/typeguard#in
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JoannaFalkowska almost 4 yearsThis is not type-safe though. If you define
response: {'a': number}
,has('items', response)
does not yield an expected compile-time error. -
quasiyoke almost 4 yearsErm... I've published the solution for cases when you're unable to control the type of arguments of the function
handle
. For example, this could be API endpoint handler -- everyone can accidentally send invalid data there. That's whyhandle
accepts the argument of typeunknown
(which means "everything, I don't know what"). Type-safety here means that you're unable (in compile time) to useresponse
without checking its type. E.g. here in line 24response.somethingElse
was used without type-check and TypeScript complains about that. -
JoannaFalkowska almost 4 yearsOh, so you meant "type-safe" in a completely different way than I thought. Cool one, thanks for the response!
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electrovir about 2 yearsI'm baffled that the built-in property checks don't already narrow types like this. I'm adding this to my default "include in all my TypeScript projects" code.
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Boris Verkhovskiy about 2 yearsCan you explain what
obj is T & Record<K, unknown>
means please? -
Boris Verkhovskiy about 2 yearsI keep getting
Type '<Whatever>' is not assignable to type 'never'
errors when I doif (!has(o, p) { o[p] = x }
when using your signature instead ofconst has = (obj: { [key: string]: any; }, prop: string) => {
which is what I had before. I have to keep doing(o[p] as Whatever) = x