Swagger declaration schema = @Schema(implementation = Map.class) 表示 Schema as String in swagger-ui

Question

我正在尝试创建 springdoc swagger 文档，我想以一种对客户来说更易读的方式来表示具有数据类型 Map<String, Object> 的请求正文。但是当我声明 @io.swagger.v3.oas.annotations.parameters.RequestBody(content = @Content(schema = @Schema(implementation = Map.class) 时，架构将以 String 的形式出现（附在下面的屏幕截图中）

方法声明

        @Operation(security = {@SecurityRequirement(name = "bearer-key")}, summary = "Create Data", operationId = "createData", description = "Create createData for the **`type`**. ")
    @ApiResponses(value = {
            @ApiResponse(responseCode = "201", description = "Data created", content = @Content(schema = @Schema(implementation = Map.class),
                    examples = {@ExampleObject(value = "{\n" +
                            "    \"id\": \"927d810c-3ac5-4584-ba58-7c11befabf54\",\n" +
                            "}")})),
            @ApiResponse(responseCode = "400", description = "BAD Request")})
    @PostMapping(value = "/data/type", produces = APPLICATION_JSON_VALUE, consumes = APPLICATION_JSON_VALUE)
    @io.swagger.v3.oas.annotations.parameters.RequestBody(content = @Content(schema = @Schema(implementation = Map.class),
            examples = {@ExampleObject(value = "{\n" +
                    "            \"label\":\"tourism\",\n" +
                    "            \"location\":\"France\"\n" +
                    "         }")}))
    ResponseEntity<Map<String, Object>> createData(@Parameter(name = "type", required = true) @PathVariable("type") String type, @Parameter(name = "request payload") @Valid @RequestBody Map<String, Object> body);

虽然Spring引导会根据方法签名自动推断类型，但对于数据类型Map并不清楚。例如，默认情况下，类型 Map 将被推断如下

但我想以一种更易于理解的方式向引用我的 API 的客户展示架构。我可以看到 Github 中有一张没有正确解决方案的已关闭票证。根据我的要求，请求主体应该是类型不可知的动态键值对，因此除了以 Map<String, Object> 形式接收请求之外别无他法。有没有人用 Map 类型实现了比创建自定义 request/response 模型更好的方法？

Answer 1

这是 springdoc-openapi 库的默认行为，以便忽略 Spring MVC 支持的其他可注入参数。

https://docs.spring.io/spring/docs/5.1.x/spring-framework-reference/web.html#mvc-ann-arguments

如果你想改变这个行为，你可以像下面这样排除它：

    SpringDocUtils.getConfig().removeRequestWrapperToIgnore(Map.class);

Answer 2

分享我解决该问题的方法，我已经针对 @io.swagger.v3.oas.annotations.parameters.RequestBody(content = @Content(schema = @Schema(implementation = Map.class) Schema is coming as String 问题做了一个解决方法。

我在 OpenAPI bean 声明中声明了一个名为 Map 的自定义模式，如下所示

new OpenAPI()
                .components(new Components()
                        .addSchemas("Map", new Schema<Map<String, Object>>().addProperties("< * >", new ObjectSchema())
                        ))
                    .....
                    .....

并在 Schema 声明中使用上述模式，如下所示

 @io.swagger.v3.oas.annotations.parameters.RequestBody(
            content = @Content(mediaType = APPLICATION_JSON_VALUE, 
                 schema = @Schema(ref = "#/components/schemas/Map"))

上面的方法可以用来代替ApiResponse也可以如下

 @io.swagger.v3.oas.annotations.responses.ApiResponse(responseCode = "200",
            content = @Content(mediaType = APPLICATION_JSON_VALUE, 
                 schema = @Schema(ref = "#/components/schemas/Map"))

注意：如果我们使用上述自定义模式方法，我们不需要更改或忽略 SpringDoc 内部使用的任何类型。

Answer 3

我有一个 API 端点，请求主体需要一个 HashMap。关于如何解决“示例值”问题的信息不多。带我去正确的地方。为了完整起见，我发布了我的解决方案，但所有功劳都归功于他。（PS：我试图赞成他的回答，但我没有足够的“积分”）

配置方面：

@Configuration
@OpenAPIDefinition
public class DocsConfiguration {
    @Bean
    public OpenAPI customOpenAPI() {
        Schema newUserSchema = new Schema<Map<String, Object>>()
                .addProperties("name",new StringSchema().example("John123"))
                .addProperties("password",new StringSchema().example("P4SSW0RD"))
                .addProperties("image",new StringSchema().example("https://robohash.org/John123.png"));

        return new OpenAPI()
                //.servers(servers)
                .info(new Info()
                        .title("Your app title")
                        .description("App description")
                        .version("1.0")
                        .license(new License().name("GNU/GPL").url("https://www.gnu.org/licenses/gpl-3.0.html"))
                )
                .components(new Components()
                        .addSchemas("NewUserBody" , newUserSchema)
                );
    }
}

控制器端：

    @io.swagger.v3.oas.annotations.parameters.RequestBody(
            content = @Content(mediaType = MediaType.APPLICATION_JSON_VALUE,
                    schema = @Schema(ref = "#/components/schemas/NewUserBody")))
    @PostMapping("/v1/users")
    public Response<User> upsertUser(@RequestBody HashMap<String,Object> user) {
         //Your code here
    }

Answer 4

我创建了一个 HashMap 扩展 class:

 @Schema(description = "Response-Object Map<String, EcoBalance).")
 public class EcoMap extends HashMap<String, EcoBalance> {
   @JsonIgnore
   @Override
   public boolean isEmpty() {
     return super.isEmpty();
   }
 }

将其用作响应对象

 @ApiResponse(responseCode = "200", content = @Content(mediaType = .., schema = @Schema(implementation = EcoMap.class)), headers = ..

注意 OpenAPI 3 生成器不会生成这样的客户端模型，但在 openapi.yml 中被正确引用（甚至验证）。

Answer 5

我想更新 for my situation. I had to combine the answer by rodiri and this answer by Ondřej Černobila to the SO question SpringDoc - How to Add schemas programmatically。我正在使用 java 11、spring-boot-starter-parent 2.5.6 和 springdoc-openapi-ui 1.5.12，我认为它使用的是 swagger 3.52 .5

<!-- https://mvnrepository.com/artifact/org.springdoc/springdoc-openapi-ui -->
<dependency>
    <groupId>org.springdoc</groupId>
    <artifactId>springdoc-openapi-ui</artifactId>
    <version>1.5.12</version>
</dependency>

我的配置

import io.swagger.v3.oas.annotations.OpenAPIDefinition;
import io.swagger.v3.oas.models.media.ObjectSchema;
import io.swagger.v3.oas.models.media.StringSchema;
import org.springdoc.core.customizers.OpenApiCustomiser;
import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.Configuration;

@Configuration
@OpenAPIDefinition
public class DocsConfiguration {

  @Bean
  public OpenApiCustomiser openApiCustomiser() {
    return openApi -> {
      var NewUserBodySchema = new ObjectSchema()
          .name("NewUserBody")
          .title("NewUserBody")
          .description("Object description")
          .addProperties("name", new StringSchema().example("John123"))
          .addProperties("password", new StringSchema().example("P4SSW0RD"))
          .addProperties("image", new StringSchema().example("https://robohash.org/John123.png"));

      var schemas = openApi.getComponents().getSchemas();
      schemas.put(NewUserBodySchema.getName(), NewUserBodySchema);
    };
  }

}

对于我的端点，我使用的是 returns 地图，因此它与接受的答案不同。

@GetMapping(value = "/{userId}")
@Operation(
    summary = "Get Something",
    description = "Some desciption",
    responses = {
        @ApiResponse(
            responseCode = "200",
            description = "The Map Response",
            content = {
                @Content(
                    mediaType = MediaType.APPLICATION_JSON_VALUE,
                    schema = @Schema(ref = "#/components/schemas/NewUserBody")
                )
            })
    }
)
public ResponseEntity<Map<String, Object>> getMap(@PathVariable String userId)

Answer 6

我今天运行自己解决了这个问题。事实证明，这实际上是 Swagger 的设计问题（@see ）。

尽管如此，我也尝试过使用此处和其他线程中的方法。

这是我的 OpenAPI，其中包含一个 Map<Integer,String>:

的自定义架构

@Configuration
@OpenAPIDefinition(
        info = @io.swagger.v3.oas.annotations.info.Info(
                title = "ACME Inc. REST API",
                version = "1.0",
                description = "This is an overview of all REST endpoints of this application",
                contact = @io.swagger.v3.oas.annotations.info.Contact(name = "John Doe", url = "https://acme-inc.com/", email = "john.doe@acme-inc.com")
        )
)
public class OpenAPIConfig {
        public static final String ERROR_CODE_MAPPER = "ErrorCode-Mapper";

        @Bean
        public OpenApiCustomiser openApiCustomiser() {
                return openApi -> {
                        Components components = openApi.getComponents();
                        for(Schema<?> schema: buildCustomSchemas()) {
                                components.addSchemas(schema.getName(), schema);
                        }
                };
        }

        private static List<Schema<?>> buildCustomSchemas() {
                ArrayList<Schema<?>> result = new ArrayList<>();

                Schema<?> integerStringMap = new Schema<Map<Integer, String>>()
                        .name(ERROR_CODE_MAPPER)
                        .type("object")
                        .addProperty("error code", new StringSchema().example("Error message belonging to the error code")).example(getErrorCodeExample());
                result.add(integerStringMap);
                // Build more custom schemas...

                return result;
        }

        private static Map<Integer, String> getErrorCodeExample() {
                Map<Integer, String> example = new HashMap<>();
                example.put(666, "Oh no..., the devil himself  showed up and stopped your request");
                return example;
        }
}

(注意： 查找您的 swagger 源代码 io.swagger.v3.oas.models.media 以获得有用的实用程序类，例如 StringSchema。您没有写一切从零开始。）

这是我的 REST 端点：

@Operation(summary = "This endpoint returns a list of system error codes, that can occur during processing requests.")
@ApiResponses(value = {
        @ApiResponse(
                responseCode = "200",
                description = "Map of all system error codes mapping to their messages",
                content = {@Content(mediaType = MediaType.APPLICATION_JSON_VALUE, schema = @Schema(ref = "#/components/schemas/"+ ERROR_CODE_MAPPER))}
        )
})
@GetMapping("/error-codes")
public Map<Integer, String> listErrorCodes() {
    // return your map here...
}

这会产生如下内容：

重要的是要知道在 JSON 对象中，键总是 string 类型。因此不必显式编写类型。考虑到这一点，这是架构：

Swagger declaration schema = @Schema(implementation = Map.class) 表示 Schema as String in swagger-ui

Swagger declaration schema = @Schema(implementation = Map.class) represents Schema as String in swagger-ui

spring-boot

swagger

springdoc

springdoc-ui