Question

我正在开发一个REST api，并试图将Swagger和Jersey一起使用我按照这里给出的指南： - https://github.com/swagger-api/swagger-core/wiki/Swagger-Core-Jersey-2.X-Project-Setup-1.5
我使用ResourceConfig代替Application类，并相应地修改了指南中的步骤。我正在grezzly上部署API。

这里是主文件： -

package com.example;

import org.glassfish.grizzly.http.server.HttpServer;
import org.glassfish.jersey.grizzly2.httpserver.GrizzlyHttpServerFactory;
import org.glassfish.jersey.server.ResourceConfig;

import java.io.IOException;
import java.net.URI;

import io.swagger.jaxrs.config.*;

/**
 * Main class.
 *
 */
public class Main {
    // Base URI the Grizzly HTTP server will listen on
    public static final String BASE_URI = "http://localhost:8080/";

    /**
     * Starts Grizzly HTTP server exposing JAX-RS resources defined in this application.
     * @return Grizzly HTTP server.
     */
    public static HttpServer startServer() {
        // create a resource config that scans for JAX-RS resources and providers
        // in com.example package
        final ResourceConfig rc = new ResourceConfig().packages("com.example");
        rc.register(io.swagger.jaxrs.listing.ApiListingResource.class);
        rc.register(io.swagger.jaxrs.listing.SwaggerSerializers.class);
        BeanConfig beanConfig = new BeanConfig();
        beanConfig.setVersion("1.0.2");
        beanConfig.setSchemes(new String[]{"http"});
        beanConfig.setHost("localhost:8080");
        beanConfig.setBasePath("resources/api");
        beanConfig.setResourcePackage("io.swagger.resources");
        beanConfig.setScan(true);

        // create and start a new instance of grizzly http server
        // exposing the Jersey application at BASE_URI
        return GrizzlyHttpServerFactory.createHttpServer(URI.create(BASE_URI), rc);
    }

    /**
     * Main method.
     * @param args
     * @throws IOException
     */
    public static void main(String[] args) throws IOException {
        final HttpServer server = startServer();
        System.out.println(String.format("Jersey app started with WADL available at "
                + "%sapplication.wadl\nHit enter to stop it...", BASE_URI));
        System.in.read();
        server.stop();
    }
}

现在，当我尝试访问网址http://localhost:8080/swagger.json上的swagger.json时，我收到500，请求失败错误。
为什么这样，我该如何调试呢？

Answer 1

看看这个项目：

https://github.com/sofien-hamdi/rest-mongo

我已使用BeanConfig配置它。

package com.kt.examples.rs.configuration;

import org.glassfish.jersey.server.ResourceConfig;
import org.glassfish.jersey.server.wadl.internal.WadlResource;
import org.springframework.context.annotation.Configuration;

import com.kt.examples.rs.controller.UserController;

import io.swagger.jaxrs.config.BeanConfig;
import io.swagger.jaxrs.listing.ApiListingResource;
import io.swagger.jaxrs.listing.SwaggerSerializers;

@Configuration
public class JerseyConfig extends ResourceConfig {

  public JerseyConfig() {
    this.registerEndpoints();
    this.configureSwagger();
  }

  private void registerEndpoints() {
    this.register(UserController.class);
    this.register(WadlResource.class);
    // http://localhost:9090/v1/application.wadl
  }

  private void configureSwagger() {
    this.register(ApiListingResource.class);
    this.register(SwaggerSerializers.class);
    BeanConfig config = new BeanConfig();
    config.setConfigId("spring-boot-jaxrs-swagger");
    config.setTitle("Spring boot jaxrs swagger integration");
    config.setVersion("v1");
    config.setBasePath("/");
    config.setResourcePackage("com.kt.examples.rs");
    config.setPrettyPrint(true);
    config.setScan(true);
    // http://localhost:9090/v1/swagger.json
  }
}

如何在泽西岛使用带有ResourceConfig的招摇？

1 个答案: