Question

I m 处理产生型号的问题。一份普通照会正向我提供两种类型之一,我希望逐一介绍。

类型:

type Add = {
    type:  add ;
    id: string;
}
  
type Remove = {
    type:  remove ;
    id: string;
}
  
type Action = Add | Remove;

type addType = Extract<Action, {type:  add }>;
    // ^? Add

然而,我的变量描述不同:

type ActionsCombined = {
    type:  add  |  remove ;
    id: string;
}

type addTypeCombined = Extract<ActionsCombined, {type:  add }>;
    // ^? never

两者似乎与我相当,但型号却看不到。是否有办法从<条码>中提取<条码>+/条码>备选案文?

https://www.typescriptlang.org/playsl=2&ssc=17&pln=3&pc=16#code/C4TwgpgBAggJnKBevAUFLVSQFxQDkAhgoQNybYCWcBAzsA7UB2A5PQL5XQBABCAFA9DgKDNh

或者,是否可以将nes变式代表制成两个无关的变量?

Answer 1

一种解决办法是将联合行动类型改为通用行动类型:

type CombinedActions<T extends  add  |  remove > = {
    type: T;
    id: string;
}

type BlankAction = CombinedActions; // ERROR: Generic type  CombinedActions  requires 1 type argument(s)
type AddAction = CombinedActions< add >; // type AddAction = { type: "add"; id: string; }
type RemoveAction = CombinedActions< remove >; // type AddAction = { type: "remove"; id: string; }
type TypoRemoveAction = CombinedActions< remive >; // ERROR: Type  "remive"  does not satisfy the constraint  "add" | "remove" .

https://www.typescriptlang.org/play?#code/C4TwgpgBAwg9gWwEYEsB2EAmBBAxsONAZwB4AVKCAD2AjU2KgHIBDTTZqAHxYCcICOADcIzAHxQAvFACIKooSAC4o5ANyLlKTBuLA+6AOY6AvPmroQ2A2rG8BItN

友情链接