我正在为Jersey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:
的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;
}
}
以下是添加的库列表:
任何人都可以帮助我吗?
答案 0 :(得分:2)
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>
答案 1 :(得分: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。通过实施您自己的Application
,您将能够指定要使用的提供商。举个例子,你可以做的是:
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);
}
}
在部署文件中添加应用程序:
<?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>
答案 2 :(得分:0)
Employee类应该实现Serializable接口
答案 3 :(得分:-1)
我能够自己解决这个问题。这是因为构建路径中包含冲突的jar。这是jar文件的快照。
答案 4 :(得分:-1)
我管理一个需要添加REST Web服务的旧项目。这个没有Maven。
对于球衣2.25(最后一次使用Java SDK 1.7编译),我解决了添加jar的问题
jersey-media-jaxb-2.25.jar