API Visualization with Scalar

Program.cs

using FastEndpoints.Swagger;
using Scalar.AspNetCore; //dotnet add package Scalar.AspNetCore

var bld = WebApplication.CreateBuilder(args);
bld.Services
   .AddFastEndpoints()
   .SwaggerDocument(); //define a swagger doc - v1 by default

var app = bld.Build();
app.UseFastEndpoints(); //must come before the UseOpenApi() call

if (app.Environment.IsDevelopment())
{
    //scalar by default looks for the swagger json file here: 
    app.UseOpenApi(c => c.Path = "/openapi/{documentName}.json");    
    app.MapScalarApiReference();
}

app.Run();

//scalar ui can be accessed at: http://localhost:{port}/scalar/v1

If I use this approach, the Model names include namespaces too, i.e. ScalartestMyRequest I tried multiple ways, but could not transform the model names, expected name MyRequest, Do you have any idea how to resolve this? @dj-nitehawk

@mkosucu
that has nothing to do with scalar. pls refer to the following docs:
schema names: https://fast-endpoints.com/docs/swagger-support#short-schema-names
endpoint names: https://fast-endpoints.com/docs/swagger-support#short-endpoint-names

I have been looking somewhere completely irrelevant! Many thanks for your help @dj-nitehawk . Appreciated.

Thanks, @dj-nitehawk! I think it's a good time to upgrade the FastEndpoints documentation and mention Scalar UI.

I create another alternative. Example

don't think that's gonna work: FastEndpoints/FastEndpoints#906 (comment)

So I have discovered that if you want the document name to be different, you must specify the document name in the MapScalarApiReference options. This is seemingly because Scalar defaults to "v1". Please let me know if there is a smoother way to do this.

Example:

const string openApiDocName = "my-api-v1";

app.UseOpenApi(options =>
{
  options.Path = "/openapi/{documentName}.json";
});
app.MapScalarApiReference(
  "/api-reference",
  options => options
             .WithTheme(ScalarTheme.Kepler)
             .AddDocument("v1", "My API v1", $"/openapi/{openApiDocName}.json"));