Question

我有一个现有的带有REST API的maven项目。我正在尝试将Swagger与项目集成。当我运行项目时，我得到了空的Swagger UI页面。没有加载API。我哪里错了？

的pom.xml

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

我使用的是自定义Application类（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 UI的index.html中（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>

Answer 1

<强> UPDATE2
我不明白为什么你使用了2 org.glassfish.jersey.servlet.ServletContainer它不起作用

确保您提到您的自定义应用程序是通过在web.xml

中提及的泽西容器加载的

如下所示

swagger-jersey2-jaxrs

依赖将足够休息将被传递。我不知道您正在使用哪种版本的球衣，但您必须延长ResourceConfig才能自定义应用程序Application课程本身

您必须在扩展资源配置中注册这两项功能

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

您的自定义应用程序类必须在您的web.xml中提及，如此

    <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>

<强>更新

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

最好的是这是swagger.json的相对路径所以它应该是这样的

 url: "swagger.json"

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

并且您在Resource Config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);
 }

请勿使用@ApplicationPath

你的Swagger.json应该在fallowing路径下可用

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

swagger ui与不显示API的maven项目集成

1 个答案: