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与maven项目的Swigger ui集成不显示API

swagger-ui swagger jersey maven java

Rubal Jain · 技术社区 · 7 年前

我有一个带有RESTAPI的现有maven项目。我正试图把招摇和这个项目结合起来。当我运行该项目时,我会得到一个空的招摇UI页面。未加载API。我哪里做错了?

pom.xml

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

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);
}
}

在索引中。Swagger UI的html(复制自文件夹到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
}

<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>

1 回复 | 直到 7 年前

harsha kumar Reddy 7 年前

我不明白你为什么用2 org.glassfish.jersey.servlet.ServletContainer 这行不通

通过在web.xml中提及,确保您提到的自定义应用程序是通过jersey容器加载的

swagger-jersey2-jaxrs

依赖性将足够了,其余部分将通过传递方式删除。我不知道你正在使用哪个版本的球衣,但你必须扩展 ResourceConfig Application 类本身

您必须在扩展资源配置中注册这两个功能

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

您的自定义应用程序类必须在web中提及。像这样的xml

    <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>

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

 url: "swagger.json"

http://localhost:8080/OpenEMPIStorage/rest/api-docs 是在这个昂首阔步复制。json文件

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

你的傲慢。json应该在fallowing path下可用

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