Could you not just have type ErrorType<T extends ApplicationError | Error> = T?
type ErrorType<T extends ApplicationError | Error> = T
Yes, we can also use something similar. In our case, T is an object having error property. If you try to use this, you will get an object (type e = {error: ServerError;}) instead of either ApplicationError or Error. You can try like,
(type e = {error: ServerError;})
type ErrorType<T extends {error: ApplicationError | Error}> = T['error']
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Could you not just have
type ErrorType<T extends ApplicationError | Error> = T?Yes, we can also use something similar. In our case, T is an object having error property. If you try to use this, you will get an object
(type e = {error: ServerError;})instead of either ApplicationError or Error. You can try like,
type ErrorType<T extends {error: ApplicationError | Error}> = T['error']