这就是诀窍：

type ActionType =
  | 'fight'
  | 'swim'
  | 'dane'

interface CreatorBase<TActions> {
  typeId: string;
  actionType:  TActions
}

interface UntrainedCreator extends CreatorBase<Omit<ActionType, 'swim'>>
{
}

interface CreatorSwim  extends CreatorBase<'swim'> {
  trained?: string;
}

export type Creator = 
| UntrainedCreator
| CreatorSwim

const swim: Creator = {actionType: 'swim', typeId : 'i', trained: 'yes'};
const  dane : Creator = {actionType: 'dane', typeId : 'i', };
const  daneError : Creator = {actionType: 'dane', typeId : 'i', trained: ''}; // error

Playground

所以在这种情况下，我会想象条件类型是你的救星。你可以检查你的动作类型是否真的是“游泳”，如果是的话，你可以在你想要添加的按键上进行Map。
请看下面的代码片段：

type ActionType = "swim" | "run" | "walk"

type Creator <T extends ActionType> = {
    actionType: T
    type: string
} & {[k in T extends "swim" ? 'trained' : never]?: any}

const x: Creator<"swim"> = {
actionType: "run",
type: "swim",
trained: true
}

type ActionType =
  | 'fight'
  | 'swim'
  | 'dane'

interface Trained {
  playerNumber: number,
  type: string
}

interface CreatorBase {
  typeId: string;
  actionType:  ActionType
}

interface UntrainedCreator extends CreatorBase
{
  trained?: never;
}

interface CreatorSwim  extends CreatorBase {
  actionType: 'swim'
  trained?: Trained;
}

export type Creator = 
| UntrainedCreator
| CreatorSwim