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找不到媒体类型= {application / xml，q = 1000}的MessageBodyWriter - Jersey Jaxb

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我正在和Jersey一起编写一个RESTful Web服务 . 我想以XML格式向消费者返回一个自定义对象 . 我得到的错误是：

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

以下是代码：

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>

Service Class

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;
    }

}

以下是添加的库列表：

Jars

谁能帮我？

rest jaxb jersey jax-rs

5 回答

  • 0

    我自己能够解决这个问题 . 这是因为构建路径中包含冲突的jar . 这是jar文件的快照 .

    enter image description here

    回复于
  • 1

    我管理一个遗留项目，我需要添加一个REST Web服务 . 这没有Maven .

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

    jersey-media-jaxb-2.25.jar

    回复于
  • -1

    我相信你的错误在web.xml中 . 尝试在web.xml中将您的部分更改为此部分 .

    <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>
    回复于
  • -1

    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序列化的提供程序 . 通过实现自己的 Application ，您将能够指定要使用的提供程序 . 举个例子，你可以做的是：

    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);
    }
}

    并在部署文件中添加应用程序：

    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>
    回复于
  • 2

    Employee类应该实现Serializable接口

    回复于

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