Is there a way to extract a particular property of a union type and create a type from that?
Let's say we have this:
type Actions =
| {
type: "ADD_COLUMN";
newColumnIndex: number;
column: SelectorColumnData;
}
| {
type: "ITEM_CHECKED";
columnIndex: number;
item: SelectorItem;
};
Is there some magic generic stuff we can use to produce something like:
type ActionTypes = GetTypes<Actions>
// ActionTypes = "ADD_COLUMN" | "ITEM_CHECKED"
You can get all possible values of a key from union of objects by just accessing a property from the type. As mentioned in the comments it can be done using :
Action["type"];
But if you really want to create a type. You will have to use extends too, so TS can let you safely access the property.
type Actions =
| {
type: "ADD_COLUMN";
newColumnIndex: number;
column: string;
}
| {
type: "ITEM_CHECKED";
columnIndex: number;
item: string;
};
type GetTypes<T extends { type : string }> = T['type'];
type ActionTypes = GetTypes<Actions>
Ok, found the answer elsewhere. Doesn't even require generics:
type ActionTypes = Actions["type"];
Actions["type"]- However, if one of the constituents ofActionsdoes not have atypeproperty, it'll be a little more complicated. You would have to extract all constituents first:Extract<Actions, { type: any }>["type"].