Say I have an array const arr = ['key1', 'key2', 'key3']. How do I make a type like
type Test = { str: Items<arr> }
so that Test types have a str attribute that is either 'key1', 'key2', 'key3'?
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2 Answers
By default TypeScript will infer arr type as string[]. If we want to infer it as ['key1', 'key2', 'key3'] we can either do as const or as ['key1', 'key2', 'key3']
To get union type of object values you need to pass a union of object indices as a type index. type ValueOf<T> = T[keyof T]. But in this case we have an array so just [number] will do.
const arr = ['key1', 'key2', 'key3'] as const;
type Test = { str: typeof arr[number] }
const test: Test = {
str: 'key1'
}
4 Comments
abc
I thought this would work but vscode shows
str's type as (property) str: string instead of the expected 'key1' | 'key2' | 'key3'
Temoncher
Did you add
as const or as ['key1', 'key2', 'key3'] at the end of the arr definition?abc
You're right, what's
as const mean here?Temoncher
as const is a const assertion which tells TypeScript to treat this variable as a readonly constant. In case of readonly constant TypeScript can safely infer the type as literal ['key1', 'key2', 'key3'] because assertion guarantees that the array will not be mutated later in the code.Define a type for just the values of str property separately. Then use it when defining your custom type Type.
Something like the following:
type Str = 'key1' | 'key2' | 'key3';
type Type = { str: Str }
2 Comments
abc
I think this is the ideal solution and type
arr: Str[] instead. Won't mark as answer since it technically doesn't answer the question. Thanks tho!
Obum
I doubt this doesn't answer your question except if you didn't properly convey what you wanted to achieve. You want
Test types to have an str attribute that is of either of those 3 values. This means Test types, technically, are objects with the str field. (they can have other fields). And if the value of the str attribute is either of those 3 values, the taking the type Str is enough, and not the type Str[] because Str[] will require an array. do you understand?