防止 Swagger 自动添加某些模型

Question

我使用Spring Boot框架构建了一个REST接口。然后，我使用Swagger版本2.9.2来生成文档。从下面的照片可以看到，Swagger自动检测了很多模型。

所以，我的问题是:我怎么告诉Swagger该曝光哪些模特？

下面是我的控制器的Swagger配置和代码片段。

@Bean
public Docket api() {
    return new Docket(DocumentationType.SWAGGER_2)
            .select()
            .apis(RequestHandlerSelectors.basePackage("my.package"))
            .paths(PathSelectors.any())
            .build()
            .apiInfo(API_INFO)
            .useDefaultResponseMessages(false);
}

控制器：

@PostMapping(value = "/import", produces = MediaType.APPLICATION_JSON_VALUE, consumes = MediaType.MULTIPART_FORM_DATA_VALUE)
public ResponseEntity<?> importData(HttpServletRequest request) {

    // processing...

    return ResponseEntity.created(uri)
        .body(new ResponseMessage(HttpStatus.CREATED, "Your data is being processed"));
}

Answer 1

Springfox Swagger2通过GET /v2/api-docs获取UI数据，这些数据会映射到spring fox . documentation . swagger 2 . web . swagger 2 controller . GET documentation()。因此，您可以创建一个bean来代替“ServiceModelToSwagger2Mapper”:

@Primary
@Component
class CustomServiceModelToSwagger2Mapper : ServiceModelToSwagger2MapperImpl() {
    @Autowired
    private lateinit var modelMapper: ModelMapper
    @Autowired
    private lateinit var parameterMapper: ParameterMapper
    @Autowired
    private lateinit var securityMapper: SecurityMapper
    @Autowired
    private lateinit var licenseMapper: LicenseMapper
    @Autowired
    private lateinit var vendorExtensionsMapper: VendorExtensionsMapper

    override fun mapDocumentation(from: Documentation?): Swagger? {
        if (from == null) {
            return null
        }

        val swagger = Swagger()

        swagger.vendorExtensions = vendorExtensionsMapper.mapExtensions(from.vendorExtensions)
        swagger.schemes = mapSchemes(from.schemes)
        swagger.paths = mapApiListings(from.apiListings)
        swagger.host = from.host
// ➡➡➡➡➡➡➡➡➡➡➡➡➡➡➡➡ here
        swagger.definitions = this.modelsFromApiListings(from.apiListings)
        swagger.securityDefinitions = securityMapper.toSecuritySchemeDefinitions(from.resourceListing)
        val info = fromResourceListingInfo(from)
        if (info != null) {
            swagger.info = mapApiInfo(info)
        }
        swagger.basePath = from.basePath
        swagger.tags = tagSetToTagList(from.tags)
        val list2 = from.consumes
        if (list2 != null) {
            swagger.consumes = ArrayList(list2)
        } else {
            swagger.consumes = null
        }
        val list3 = from.produces
        if (list3 != null) {
            swagger.produces = ArrayList(list3)
        } else {
            swagger.produces = null
        }

        return swagger
    }

    private fun fromResourceListingInfo(documentation: Documentation?): ApiInfo? {
        if (documentation == null) {
            return null
        }
        val resourceListing = documentation.resourceListing ?: return null
        return resourceListing.info ?: return null
    }


    /**
     * @see ModelMapper
     */
    internal fun modelsFromApiListings(apiListings: Multimap<String, ApiListing>): Map<String, Model>? {
        val definitions = newTreeMap<String, springfox.documentation.schema.Model>()
        for (each in apiListings.values()) {
// ➡➡➡➡➡➡➡➡➡➡➡➡➡➡➡➡ here
            // definitions.putAll(each.models)
            definitions.putAll(each.models.filter {
                it.value.qualifiedType.startsWith("com.cpvsn")
                        && it.value.type.typeBindings.isEmpty    
            })
        }
        return modelMapper.mapModels(definitions)
    }
}

Answer 2

@Api、@ApiOperation、@ApiResponses、@ApiParam、@ApiIgnore、@ApiModel、@ApiModelProperty等

Answer 3

您可以使用：

@Bean
public Docket api() {
    return new Docket(DocumentationType.SWAGGER_2).select()
            .apis(RequestHandlerSelectors.basePackage("com.package"))
            .paths(PathSelectors.regex("/api.*")).build().apiInfo(apiInfo())
            .ignoredParameterTypes(Timestamp.class);
}

这对我很有效。在ignoredParameterTypes中指定类名后，它不再出现在swagger ui中。