TypeScript - How can omit some items from an Enum in TypeScript?

Question

I defined an Enum to clarify APIs request status:

const enum Errcode{     Ok=0,     Error=1,     AccessDeny=201,     PostsNotFound=202,     TagNotFound=203,     //... }  type SuccessErrcode =Errcode.Ok; type NotFoundError =Errcode.PostsNotFound|Errcode.TagNotFound; type ErrorErrcode=/* there */; 

How can I define the ErrorErrcode that means all items of Errcode except Errcode.Ok (and it should include all items of NotFoundError)?

I can't define the more granular types and Union them likes this:

enum SuccessErrcode {     Ok =0, } enum NotFoundErrcode {     PostsNotFound=202,     TagNotFound=203, } enum ErrorErrcode {     Error=1, } type Errcode =SuccessErrcode|NotFoundError|SuccessErrcode; 

If I do this, I will can't use Errcode.xxx - for use a code, I must to know where it be assigned of.(e.g. from Errcode.TagNotFound to NotFoundError.TagNotFound). And consider that - when there have TagErrcode and NotFoundErrcode, the TagNotFound=203 will be defined twice.

Answer 1

As of TypeScript 2.8 and the addition of conditional types, you can use the built-in Exclude to exclude certain enum values:

const enum Errcode {     Ok=0,     Error=1,     AccessDeny=201,     PostsNotFound=202,     TagNotFound=203,     //... }  type SuccessErrcode = Errcode.Ok; type NotFoundError = Errcode.PostsNotFound|Errcode.TagNotFound; type ErrorErrcode = Exclude<Errcode, Errcode.Ok>; 

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Answer 2

You would first define the more granular types. Perhaps something like this:

enum ErrorCode {     Error = 1,     AccessDeny = 201,     PostsNotFound = 202,     TagNotFound = 203, }  enum SuccessCode {     Ok = 0 } 

You can then define a Union Type to be either a SuccessCode or a ErrorCode:

type ResultCode = ErrorCode | SuccessCode; 

which you can then use like this:

const myResult1: ResultCode = ErrorCode.AccessDeny; const myResult2: ResultCode = SuccessCode.Ok;