connecting express middleware with own routes

[HardCoreQual]

I trying to connect swagger-ui-express

in express it is connected with code

import swaggerUi from 'swagger-ui-express';
app.use('/swagger', swaggerUi.serve, swaggerUi.setup(scheme, options));

here I connect it using

export const routing: Routing = {
  v1: apiRouter,
  internal: internalRouting,
};

const options = {
  swaggerOptions: {
    validatorUrl: null,
  },
};

const schema = JSON.parse(new OpenAPI({
  routing, // the same routing and config that you use to start the server
  config: serverConfig,
  version: '1.0.0',
  title: 'Some title',
  serverUrl: 'https://localhost:9999/image',
}).getSpecAsJson());


export const swagger = appFactory
  .addExpressMiddleware(swaggerUi.serve[0] as any)
  .addExpressMiddleware(swaggerUi.serve[1] as any)
  .addExpressMiddleware(swaggerUi.setup(schema as any, options))
  .build({
    method: 'get',
    input: z.object({}),
    output: z.object({}),
    async handler() {
      return {};
    },
  });


routing.swagger = swagger;

but seems like it permit to make request only to route that exist in routing, but don't for routes that will added by swaggerUI

also try it with express router

export const swagger = appFactory
  .addExpressMiddleware(express.Router().use('/swagger', swaggerUi.serve, swaggerUi.setup(schema, options)))
  .build({
    method: 'get',
    input: z.object({}),
    output: z.object({}),
    async handler() {
      return {};
    },
  });

but it don't help

Of course here can just create express app and attachRouting

[RobinTail]

Wow. Thank you @HardCoreQual for showing this.

Instead all of that, I recommend to separate the actual routes from this UI by using attachRouting().
This approach is described here:
https://github.com/RobinTail/express-zod-api#connect-to-your-own-express-app

This way you wouldn't need to create fake endpoint.