MessageBodyWriter 未找到媒体类型={application/xml, q=1000} - Jersey + Jaxb

Question

我正在用 Jersey 编写一个 RESTful Web 服务。我想以 XML 形式将自定义对象返回给消费者。我得到的错误是：

找不到媒体类型={application/xml, q=1000}，类型=类 com.test.ws.Employee，genericType=类 com.test.ws.Employee 的 MessageBodyWriter。

下面是代码：

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com/xml/ns/javaee"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
id="WebApp_ID" version="3.0">
<display-name>com.vogella.jersey.first</display-name>
<servlet>
    <servlet-name>Jersey REST Service</servlet-name>
    <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
    <!-- Register resources and providers under com.vogella.jersey.first package. -->
    <init-param>
        <param-name>jersey.config.server.provider.packages</param-name>
        <param-value>com.test.ws</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> 

服务类

package com.test.ws;

@Path("/hello")
public class Hello {

    @GET 
    @Path("/sayHello")
    @Produces(MediaType.APPLICATION_XML)
    public Employee sayHello() {
        Employee employee = new Employee();
        employee.setEmpId(1);
        employee.setFirstName("Aniket");
        employee.setLastName("Khadke");
        return employee;
    }
}

Employee.java

package com.test.ws;

import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement(name = "employee")
public class Employee {

    public String firstName;

    public String lastName;
    public int empId;

    public Employee(String firstName, String lastName, int empId) {
        super();
        this.firstName = firstName;
        this.lastName = lastName;
        this.empId = empId;
    }

    public Employee() {
        super();
    }

    @XmlElement
    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    @XmlElement
    public String getLastName() {
        return lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    @XmlElement
    public int getEmpId() {
        return empId;
    }

    public void setEmpId(int empId) {
        this.empId = empId;
    }

}

这里是添加的库列表：

谁能帮帮我？

Answer 1

我管理一个需要添加 REST Web 服务的遗留项目。这没有 Maven。

对于 jersey 2.25，最后使用 Java SDK 1.7 编译，我解决了添加 jar

jersey-media-jaxb-2.25.jar

Answer 2

Employee类应该实现Serializable接口

Answer 3

解决问题的一种方法是创建自定义javax.ws.rs.core.Application 或org.glassfish.jersey.server.ResourceConfig。您的服务器似乎没有检测到您的序列化提供程序。通过实现您自己的Application，您将能够指定要使用的提供程序。对于您的示例，您可以做的是：

MyApplication.java

package com.test.ws;

public class MyApplication extends ResourceConfig {
    public MyApplication() {
        //register your resources
        packages("com.test.ws");
        //if you're using Jackson as your XMLProvider for example
        register(JacksonJaxbXMLProvider.class);
    }
}

并将应用程序添加到您的部署文件中：

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xmlns="http://java.sun.com/xml/ns/javaee"
         xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
         id="WebApp_ID" version="3.0">
    <display-name>com.vogella.jersey.first</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.test.ws.MyApplication</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>

Answer 4

我能够自己解决问题。这是因为构建路径中包含有冲突的 jar。这是 jar 文件的快照。

Answer 5

我相信您的错误在 web.xml 中。尝试在您的 web.xml 中将您的部分更改为此。

<servlet>
       <servlet-name>Jersey REST Service</servlet-name>
       <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
       <!-- Register resources and providers under com.vogella.jersey.first package. -->
       <init-param>
           <param-name>com.sun.jersey.config.property.packages</param-name>
        <param-value>com.test.ws</param-value>
       </init-param>
       <load-on-startup>1</load-on-startup>
</servlet>