How to initialize swagger from openapi resource file

Question

I'm using kotlin spring boot web mvc application. I've created an api-first file openapi.yml that contains api description. Also I've added required dependencies, like:

1. Gradle openapi generator plugin id("org.openapi.generator") version "6.3.0"
2. Openapi implementation("org.springdoc:springdoc-openapi-ui:${libs.versions.openapi.get()}")

I've successfully generated apis / model with gradle generator from my file, but I need to make it as a template for swagger. I've reviewed all the springfoc's settings and it seems, that I can only forward a generated documentation with a springdoc.swagger-ui.url property.

Is there any way to initialized swagger from a openapi.yml file manually?

Answer 1

I've managed to find how to resolve this issue.

For springdoc the right way is described here: https://springdoc.org/faq.html#_how_can_use_custom_jsonyml_file_instead_of_generated_one

My steps:

1. Add springdoc settings:

springdoc:
  api-docs:
    enabled: false
  swagger-ui:
    url: /openapi.yaml

2. As it's written there - it's very important to place yaml file into proper location: src/main/resources/static
3. For multi-module projects api docs can be placed at root projects and then copied into mentioned above path with a custom gradle task, like:

register<Copy>("copyOpenApiDoc") {
        from(rootProjectOpenapiFolder.resolve("openapi.yaml"))
        into(buildDir.resolve("resources/main/static"))
}

Perhaps not a beautiful solution, but it works