Swagger 无法为 HTTP“PATCH”生成文档

Question

我已遵循以下
博客条目： http://kingsfleet.blogspot.co.uk/2014/02/transparent-patch-support-in-jax-rs-20.html
https://github.com/jersey/jersey/tree/2.6/examples/http-patch
在 Jersey 2.6 中创建支持 HTTP“PATCH”方法的端点 依赖版本：

-Jersey: 2.6
-swagger-jersey2-jaxrs_2.10: 1.3.12

问题？ 为什么 Patch 端点没有被列为 swagger ui 文档的一部分？

分析：

如果我使用此注释进行注释，则会生成该端点的文档，但没有交互。

@com.wordnik.swagger.jaxrs.PATCH

配置

JerssyApplicationInitializer

packages(true, "com.test.account.endpoint", "com.wordnik.swagger.jaxrs.json");
        //Swagger Configuration
        register(new ApiListingResourceJSON(), 10);
        register(JerseyApiDeclarationProvider.class);
        register(JerseyResourceListingProvider.class);

        //Genson Converter
        register(GensonJsonConverter.class, 1);
        register(createMoxyJsonResolver());

我不确定，如果我遗漏了什么，任何帮助或指南都会有所帮助。

补丁方法文档：

 public static final String PATCH_MEDIA_TYPE = "application/json-patch+json";
    @PATCH
        //@com.wordnik.swagger.jaxrs.PATCH
        @PreAuthorize(userAuthenticationRequire=true)
        @Consumes(PATCH_MEDIA_TYPE)
        @Path("{id: .\\d+}")
        @ApiOperation(value = "Update Client Details in UIM System."
                    , response = State.class
                    , notes="Requesting User, should be the owner of the Client."
                    , consumes = PATCH_MEDIA_TYPE)
        @ApiResponses({
            @ApiResponse(code = _401, message = "If the access token is invalid.", response = String.class),
            @ApiResponse(code = _498, message = "If the access token is expired.", response = String.class),
            @ApiResponse(code = _420, message = "If Provided Input is not valid according to requirment specification."),
            @ApiResponse(code = _404, message = "If no client/app Found."),
            @ApiResponse(code = _200, message = "If Client Account has been Updated successfully. ", response=String.class)
        })
        public State updateClientDetails(@ApiParam(value="Client Id to be Updated.", required=true) @PathParam(CLIENT_ID) String clientId
                , @ApiParam(value = "Updated field and Value.", required = true) final State newState){
            //LOG.info("[ENTRY]- Received requst for updating Client {} from System.", clientId);
            System.out.println("----->" + someBean.test());
            //LOG.info("[EXIT]- Client Id {} Updation has been completed.", clientId);
            Test t = new Test();
            t.name = "Hello Test";
            System.out.println(t.name);
            return newState;
        }

Answer 1

看看你的 index.html。控制哪些 HTTP 操作是交互式的——通过将其更改为：

  window.swaggerUi = new SwaggerUi({
    url: url,
    dom_id: "swagger-ui-container",
    supportedSubmitMethods: ['get', 'post', 'put', 'delete', 'patch'],

您将在 PATCH 方法上进行交互：