相关疑难解决方法(0)

我正在尝试使用Jersey2构建示例REST服务:

这是我的POM依赖项:

<dependencies>
    <dependency>
      <groupId>junit</groupId>
      <artifactId>junit</artifactId>
      <version>3.8.1</version>
      <scope>test</scope>
    </dependency>

    <dependency>
        <groupId>org.glassfish.jersey.core</groupId>
        <artifactId>jersey-client</artifactId>
        <version>2.4.1</version>
    </dependency>
          <dependency>
        <groupId>org.glassfish.jersey.containers</groupId>
        <artifactId>jersey-container-servlet</artifactId>
        <version>2.6</version>
    </dependency>
    <dependency>
        <groupId>org.glassfish.jersey.core</groupId>
        <artifactId>jersey-server</artifactId>
        <version>2.6</version>
    </dependency>
    <dependency>
      <groupId>org.glassfish.jersey.bundles.repackaged</groupId>
      <artifactId>jersey-guava</artifactId>
      <version>2.6</version>
    </dependency>
  </dependencies>

这是我的web.xml:

<web-app>
  <display-name>jerseysample</display-name>
    <servlet>
        <servlet-name>Jersey REST Service</servlet-name>
        <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
        <init-param>
            <param-name>javax.ws.rs.Application</param-name>
            <param-value>com.adaequare.rest.config.JerseyResourceInitializer</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>Jersey REST Service</servlet-name>
        <url-pattern>/rest/*</url-pattern>
    </servlet-mapping>
</web-app>

这是我的ResourceConfig实现:

package com.adaequare.rest.config;

import org.glassfish.jersey.server.ResourceConfig;

public class JerseyResourceInitializer extends ResourceConfig {
    public JerseyResourceInitializer() {
        packages(true, "com.adaequare.resource");
    }
}

最后这是我的示例类:

package com.adaequare.resource;

import javax.ws.rs.Consumes;
import javax.ws.rs.GET;
import javax.ws.rs.PUT;
import javax.ws.rs.Path; …