typescript：基于同一对象属性的推断函数参数

Question

any

，其中应为同一对象中定义的ZOD模式的形状。这可能吗？这是一个最小的例子：

import { ComponentType } from "react";
import { z } from "zod";

type Step<
  TKey = string,
  TSchema extends z.ZodType = z.ZodType,
  TProps = unknown
> = {
  component: ComponentType<TProps>;
  schema: TSchema;
  getNextStep?: (data: z.infer<TSchema>) => TKey;
};

type Config<TConfig> = {
  [K in keyof TConfig]: TConfig[K] extends Step ? TConfig[K] : unknown;
};

function createConfig<TConfig>(config: Config<TConfig>) {
  return config;
}

const config = createConfig({
  Step1: {
    component: ({ orgId }: { orgId: string }) => null,
    schema: z.object({ userId: z.string() }),
    getNextStep: (d) => "Step2", // <---- `d` is inferred as `any`
  },
  Step2: {
    component: ({ userId }: { userId: string }) => null,
    schema: z.unknown(),
    getNextStep: () => "Step1",
  },
});

您可以尝试更简单的东西：

import type { ComponentType } from "react";
import { z } from "zod";

type Step<
    TKey = string,
    TSchema extends z.ZodType = z.ZodType,
    TProps = unknown
> = {
    component: ComponentType<TProps>;
    schema: TSchema;
    getNextStep?: (data: TSchema) => TKey;
};

const step1Schema = z.object({ userId: z.string() });
const step2Schema = z.unknown();

type Config = {
    Step1: Step<"Step1", typeof step1Schema, { orgId: string }>,
    Step2: Step<"Step2", typeof step2Schema, { orgId: string }>,
}

const config: Config = {
    Step1: {
        component: ({ orgId }) => null,
        schema: step1Schema,
        getNextStep: (data) => "Step1"
    },
    Step2: {
        component: ({ orgId }) => null,
        schema: step2Schema,
        getNextStep: (data) => "Step2"
    }
}

Answer 1