Dynamically set registration id from query param - spring-security

Question

I have the following controller to get registration ClientA when the endpoint is called.
#GetMapping("/token")
fun token(
#RegisteredOAuth2AuthorizedClient("clientA") authorizedClient: OAuth2AuthorizedClient
): ResponseEntity<String> {
val token = tokenService.getToken()
return ResponseEntity(token, HttpStatus.OK)
}
I want to have the client as a query param and dynamically start the OAuth2 process. How could I achieve it? sth like the following:
#GetMapping("/token?client={client}")
fun token(
#RegisteredOAuth2AuthorizedClient(${client}) authorizedClient: OAuth2AuthorizedClient
): ResponseEntity<String> {
val token = tokenService.getToken()
return ResponseEntity(token, HttpStatus.OK)
}

Answer 1

My solution is to use the default authorization uri and defaultSuccessUrl("/token")
The goal is to scale the client easily, and this solution can achieve the same.
Solution:
Register clientA in application.yml
clientA:
client-id: Any
redirect-uri: http://localhost/index
provider: clientA-provider
scope: launch
client-name: clientA
client-authentication-method: none
authorization-grant-type: authorization_code
Now the endpoint /oauth2/authorization/clientA is created automatically by spring security. For getting a token, call the above url.
After successful authorization, the endpoint goes to /token, due to defaultSuccessUrl("/token").
The following shows the related code snippet:
override fun configure(http: HttpSecurity) {
http.csrf()
.disable()
.oauth2Login()
.authorizationEndpoint()
.and()
.tokenEndpoint()
.and()
.defaultSuccessUrl("/token")
#Controller
#RequestMapping("/token")
class LaunchController(private val tokenService: TokenService) {
#GetMapping
fun token(): ResponseEntity<String> {
val token = tokenService.getToken()
return ResponseEntity.ok(token)
}
}