是否可以不对 output 做运行时校验，只做 typescript 检查

[tbase]

有时候输出太大，做运行时校验会很慢。但是 typescript 检查时必须的

[tbase]

可以使用 z.custom<Type>()

[lorefnon]

Translated (en):

Is it possible to skip runtime validation of the output and only perform TypeScript checks? #1
Sometimes the output is too large, making runtime validation very slow. However, this is essential for TypeScript checks.
You can use z.custom()

Yes, z.custom can be used.

You can also skip zod integration completely for one or more services.

In shared file:

// Define a plain ts interface instead of inferring it from a zod schema:
interface MyService {
 // ... 
}

// On server side:

import express from "express";
import { rpcHandler } from "@lorefnon/ts-json-rpc/lib/express";

// Create service class that implements service interface
class MyServiceImpl implements MyService { ... }

const app = express();
app.use(express.json());
app.post("/api", rpcHandler(MyServiceImpl));
app.listen(3000);

// On client side:

import { rpcClient, HttpPostTransport } from "@lorefnon/ts-json-rpc/lib/client";

// Import the type
import type { MyService } from "../shared";

// Create a typed client:
const client = rpcClient<MyService>({
  transport: new HttpPostTransport({
    url: "http://localhost:3000/api"
  })
});

// Call a remote method:
console.log(await client.hello("world"));