How do i display only the url without the UI path?

[afidegnum]

With swagger-ui I would like to return only the openapi schema without the UI,

the following code still returns the ui interface. SwaggerUi::new("").url("/api-docs/openapi.json", openapi.clone()),

[juhaku]

For such case you need to serve it through an enpoint as you would do with any other content. And also with that you do not need the utoipa-swagger-ui library to be there.

async fn get_api_doc(....) -> Json<utoipa::openapi::OpenApi> {
    let doc = // .... load the doc somewhere or from fn arguments
    Json(doc)
}

[afidegnum]

Can you please give an example? i'm a bit confused.

[juhaku]

Sure,

For example if you take a look at the warp example you can find that the apidoc.json is served via regular http endpoint handler (whatever handles the incoming requests). https://github.com/juhaku/utoipa/blob/489ddd663051a578e9285f5f4b5d87c0f626d6c0/examples/todo-warp/src/main.rs#L46-L48

Or with actix_web same example might look something like this:

#[get("/api-docs/openapi.json")]
async fn get_api_doc(apidoc: Data<OpenApi>) -> impl Responder {
    let doc = apidoc.lock().unwrap();
    HttpResponse::Ok().json(doc)
}

If you want to return only the openapi schema then you need to create a custom endpoint for it according to the framework of your choise. You cannot use the SwaggerUi type without exposing the Swagger UI to the public.

Does this help? Or have I understood your question totally wrong?

[afidegnum]

Thanks a lot.

[juhaku]

Closed due inactivity.