尝试使用satisfies RouteRecordRaw[]而不是as const。
你会得到相同的类型扣除，但ts不会因为额外的readonly s而对你大喊大叫

这里有一种方法可以确保满足所有类型的要求，同时还可以获得findRouteByName的智能感知。
router/index.ts个

import {
  RouteRecordRaw,
  _RouteRecordBase,
  createRouter,
  createWebHistory,
} from "vue-router";

type ForceMergeTypes<T, K> = T | (K & Record<never, never>);

type Paths = "/" | "somePath" | "somePath2";
type Names = "Root" | "somePath" | "somePath2";

type ExtendedRouteRecordBase = _RouteRecordBase & {
  path: ForceMergeTypes<Paths, RouteRecordRaw["path"]>;
  name?: ForceMergeTypes<Names, RouteRecordRaw["name"]>;
};

type RoutesRecord = ExtendedRouteRecordBase & RouteRecordRaw;

const routes: Array<RoutesRecord> = [
  {
    path: "/",
    name: "Root",
    component: null,
    children: [],
  },
  {
    path: "/somePath",
    name: "somePath1",
    component: null,
    children: [],
  },
  {
    path: "/somePath2",
    name: "somePath2",
    component: null,
    children: [],
  },
];

export function findRouteByName(name: (typeof routes)[number]["name"]) {
  return routes.find((route) => route.name === name);
}

const router = createRouter({
  history: createWebHistory(process.env.BASE_URL),
  routes,
});

findRouteByName("");

export default router;

字符串
使用这种方法，您可以为path和name提供任何字符串，并且还可以为添加到Path和Names类型的字符串获得intellisense。让我知道如果我错过了什么或者如果不符合您的要求。

的数据

**注意：**您可能无法在Stackblitz中获得智能感知，但是，您可以在IDE中获得它，如VSCode。这是因为Stackblitz的限制。