swagger ui 与不显示 API 的 Maven 项目集成

swagger ui integration with maven project not displaying APIs

我有一个带有 REST API 的现有 Maven 项目。我正在尝试将 Swagger 与项目集成。当我 运行 项目时,我得到空的 Swagger UI 页面。没有加载 API。我哪里错了?

pom.xml

<!-- Swagger -->
    <dependency>
        <groupId>io.swagger</groupId>
        <artifactId>swagger-jersey2-jaxrs</artifactId>
        <version>1.5.16</version>
    </dependency>

我正在使用自定义应用程序 class(不是 web.xml)

public class MyRestApplication extends ResourceConfig {
public MyRestApplication() {
    System.out.println("Entering Custom Application");
    property(ServerProperties.BV_SEND_ERROR_IN_RESPONSE, true);    
    property(ServerProperties.
    BV_DISABLE_VALIDATE_ON_EXECUTABLE_OVERRIDE_CHECK, true);
    register(RolesAllowedDynamicFeature.class);        
    register(StorageResource.class);
    register(io.swagger.jaxrs.listing.ApiListingResource.class);
    register(io.swagger.jaxrs.listing.SwaggerSerializers.class);        
}
}    

在我的 StorageResource 文件中:

@Api(value = "/Person")
@Path("")
public class StorageResource {
static Logger logger = 
Logger.getLogger(StorageResource.class.getName());
@PUT
@Path("/person-manager-resource/addPerson")
@Produces("application/json")
@Consumes("application/json")
@ApiOperation(
        value = "method api",
        notes = "method api notes"
)
public Object addPerson(String reqBody) {
    AddPerson add = new AddPerson();
    return add.addPerson(reqBody);
}
}

在Swagger的index.html中UI(从dist文件夹复制到webapp文件夹)

window.onload = function() {  
// Build a system
const ui = SwaggerUIBundle({
url: "http://localhost:8080/demographics/dgs/swagger.json",
dom_id: '#swagger-ui',
presets: [
  SwaggerUIBundle.presets.apis,
  SwaggerUIStandalonePreset
],
plugins: [
  SwaggerUIBundle.plugins.DownloadUrl
],
layout: "StandaloneLayout"
})
window.ui = ui
}

我的 web.xml 文件:

<servlet>
    <servlet-name>Jersey Web Application</servlet-name>
    <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
    <init-param>
        <param-name>jersey.config.server.provider.packages</param-name>
        <param-value>in.healthelife.DGS</param-value>
    </init-param>
    <init-param>
        <param-name>javax.ws.rs.Application</param-name>
        <param-value>in.healthelife.DGS.data.MyRestApplication</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>Jersey Web Application</servlet-name>
    <url-pattern>/dgs/*</url-pattern>
</servlet-mapping>

update2
我不明白为什么你用了 2 org.glassfish.jersey.servlet.ServletContainer 它不起作用

通过在 web.xml

中提及,确保您提到您的自定义应用程序是通过 jersey 容器加载的

如下图


swagger-jersey2-jaxrs

dependency 就足够了,rest 会被 transitively pull。我不知道您使用的是哪个版本的球衣,但您必须扩展 ResourceConfig 才能自定义应用程序而不是 Application class 本身

您必须在扩展资源配置中注册这 2 个功能

io.swagger.jaxrs.listing.ApiListingResource,
     io.swagger.jaxrs.listing.SwaggerSerializers

并且您的自定义应用程序 class 必须在您的 web.xml 中提及

    <servlet>
    <servlet-name>Jersey Web Application</servlet-name>
    <servlet-class>org.glassfish.jersey.servlet.ServletContainer
   </servlet-class>
    <init-param>
      <param-name>javax.ws.rs.Application</param-name>
      <param-value>com.foo.app.CustomApplication
     </param-value>
    </init-param>
    </servlet>

更新

url: "http://localhost:8080/OpenEMPIStorage/rest/api-docs",

最好是swagger.json的相对路径所以应该是这样

 url: "swagger.json"

并确保您在路径 http://localhost:8080/OpenEMPIStorage/rest/api-docs 中进入的 json 已复制到此 swagger.json 文件

并且您在资源配置中的代码应该像

public MyRestApplication() 
{

    System.out.println("Entering Custom Application");

    property(ServerProperties.BV_SEND_ERROR_IN_RESPONSE, true);
    // @ValidateOnExecution annotations on subclasses won't cause errors.
    property(ServerProperties.BV_DISABLE_VALIDATE_ON_EXECUTABLE_OVERRIDE_CHECK, true);
    register(RolesAllowedDynamicFeature.class);
    register(AuthenticationFilter.class);
    register(AuthenticationResponseFilter.class);
 }

不要使用 @ApplicationPath

你的 Swagger.json 应该在休闲路径

下可用
http://<ip address:portname or domain name >/<yourapp-path>/dgs/swagger.json