Creating type based on object values in an array JSON in TypeScript

Question

I have a JSON file that has a format something like this:

[
    {
        "foo1": "bar1",
        "something": "",
        "else": ""
    },
    {
        "foo2": "bar2",
        "something": "",
        "else": ""
    },
    ...
]

and I want to create a type with only the "bar" values i.e. the values of the foos of all the objects in the list.

I want to create a type that produces the following:

type bars = "bar1" | "bar2" | "bar3" | //...

but I have no idea where to even start. How do I traverse through that list and only return the values, then store them in a type?

Edit: I made a mistake. The JSON file actually looks more like this:

[
    {
        "foo": "bar1",
        "something": "",
        "else": ""
    },
    {
        "foo": "bar2",
        "something": "",
        "else": ""
    },
    ...
]

From which keys are you supposed to infer the union type from? — Terry
– Terry, Commented Feb 8, 2023 at 0:06
@Terry I'm not sure what you mean. Could you elaborate, or perhaps use simpler language? I'm not very advanced at Typescript yet. — harbar20
– harbar20, Commented Feb 8, 2023 at 0:15
In your original question you are trying to infer a union type from different keys "foo1" and "foo2", but I can see you've updated your question. With the same key "foo" it's much easier and simpler to get the union type you want. — Terry
– Terry, Commented Feb 8, 2023 at 0:23

Terry · Accepted Answer · 2023-02-08 00:29:50Z

1

If you can define your data as const then you can easily infer the union type of the foo key using mapped types:

const data = [
    {
        "foo": "bar1",
        "something": "",
        "else": ""
    },
    {
        "foo": "bar2",
        "something": "",
        "else": ""
    },
] as const;

// Expected: type bars = "bar1" | "bar2"
type bars = typeof data[number]['foo'];

See example on TypeScript playground.

edited Feb 8, 2023 at 0:29

answered Feb 8, 2023 at 0:21

Terry

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