严重：找不到 Java 类 java.util.ArrayList 和 MIME 媒体类型 application/json 的消息正文编写器

rest java jdk1.5 jersey oc4j

2022-09-03 03:37:28

我正在测试RESTful服务，当我执行时，我得到了异常，尽管我的类路径（WEB-INF / lib）中有以下jars，我没有使用Maven，我的JDK版本是1.5。有关此问题的其他问题无助于解决问题。

代码片段

@GET
@Produces("application/json")    
//@Produces({MediaType.APPLICATION_JSON}) tried this, didn't work either
public List<Emp> getEmployees() {        
    List<Emp> empList = myDAO.getAllEmployees();
    log.info("size   " + empList.size());
    return empList;
}

@XmlRootElement
public class Emp {
......

网.xml

<servlet>
    <servlet-name>Jersey Web Application</servlet-name>
    <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
    <init-param>
        <param-name>com.sun.jersey.config.property.packages</param-name>
        <param-value>test.employees</param-value>
    </init-param>
    <init-param>
        <param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
        <param-value>true</param-value>
    </init-param>
</servlet>

 <servlet-mapping>
    <servlet-name>Jersey Web Application</servlet-name>
    <url-pattern>/rest/*</url-pattern>
</servlet-mapping>

罐子列表

jersey-server-1.2.jar
jersey-core-1.2.jar
jsr311-api-1.1.jar
asm-3.1.jar
jaxb-api-2.0.jar
jaxb-impl-2.0.jar
jackson-xc-1.2.0.jar
jackson-jaxrs-1.2.0.jar
jackson-mapper-asl-1.2.0.jar
jackson-core-asl-1.2.0.jar
jettison-1.2.jar
jersey-client-1.2.jar
jersey-servlet-1.10.jar
jersey-json-1.8.jar

异常堆栈

 SEVERE: A message body writer for Java class java.util.ArrayList,
 and Java type java.util.List<test.Emp>, 
 and MIME media type application/json was not found
Nov 21, 2013 11:47:26 AM com.sun.jersey.spi.container.ContainerResponse traceException
SEVERE: Mapped exception to response: 500 (Internal Server Error)

javax.ws.rs.WebApplicationException
    at javax.ws.rs.WebApplicationException.<init>(WebApplicationException.java:97)
    at javax.ws.rs.WebApplicationException.<init>(WebApplicationException.java:55)
    at com.sun.jersey.spi.container.ContainerResponse.write(ContainerResponse.java:267)
    at com.sun.jersey.server.impl.application.WebApplicationImpl._handleRequest(WebApplicationImpl.java:1035)
    at com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:947)
    at com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:939)
    at com.sun.jersey.spi.container.servlet.WebComponent.service(WebComponent.java:399)
    at com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:478)
    at com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:663)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:856)
    at com.evermind.server.http.ServletRequestDispatcher.invoke(ServletRequestDispatcher.java:719)
    at com.evermind.server.http.ServletRequestDispatcher.forwardInternal(ServletRequestDispatcher.java:376)
    at com.evermind.server.http.HttpRequestHandler.doProcessRequest(HttpRequestHandler.java:870)
    at com.evermind.server.http.HttpRequestHandler.processRequest(HttpRequestHandler.java:451)
    at com.evermind.server.http.HttpRequestHandler.serveOneRequest(HttpRequestHandler.java:218)
    at com.evermind.server.http.HttpRequestHandler.run(HttpRequestHandler.java:119)
    at com.evermind.server.http.HttpRequestHandler.run(HttpRequestHandler.java:112)
    at oracle.oc4j.network.ServerSocketReadHandler$SafeRunnable.run(ServerSocketReadHandler.java:260)
    at oracle.oc4j.network.ServerSocketAcceptHandler.procClientSocket(ServerSocketAcceptHandler.java:230)
    at oracle.oc4j.network.ServerSocketAcceptHandler.access$800(ServerSocketAcceptHandler.java:33)
    at oracle.oc4j.network.ServerSocketAcceptHandler$AcceptHandlerHorse.run(ServerSocketAcceptHandler.java:831)
    at com.evermind.util.ReleasableResourcePooledExecutor$MyWorker.run(ReleasableResourcePooledExecutor.java:303)
    at java.lang.Thread.run(Thread.java:595)

如何解决此问题？

答案 1

问题可能出在您尝试返回结果的方式上。我也见过其他人以这种方式编写他们的服务层代码，但是泽西岛提供了一种干净利落地做到这一点的方法，它将支持JSON，XML和HTML输出，您只需要使用@Produces注释进行指定即可。这就是我所做的：

import javax.ws.rs.GET;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.GenericEntity;
import javax.ws.rs.core.Response;

@GET
@Produces( MediaType.APPLICATION_JSON )
public Response getEmployees()
{        
    List< Emp >                  matched;
    GenericEntity< List< Emp > > entity;

    matched = myDAO.getAllEmployees();
    entity  = new GenericEntity< List< Emp > >( matched ) { };

    return Response.ok( entity ).build();
}

我正在使用以下泽西岛库：

球衣核心-1.8.jar
泽西-json-1.8.jar
泽西-服务器-1.8.jar

答案 2

不能将响应定义为，因为无法识别 over 或类定义。XmlList<Emp>JAXB@XmlRootElementjava.util.Listjava.util.ArrayList

理想情况下，应为子元素集合提供一个父/根元素。

再创建一个类，以包含对象集合，如下所示，并尝试一下。EmployeesEmp

@GET
@Produces("application/json")    
public Employees getEmployees() {        
    List<Emp> empList = myDAO.getAllEmployees();
    log.info("size   " + empList.size());
    Employees employees = new Employees();
    employees.setEmployeeList(empList);

    return employees;
}

@XmlRootElement(name = "Employees")
public class Employees {

    List<Emp> employeeList;

    //setters and getters goes here
}

@XmlRootElement()
class Emp {
   //fields here
}

请尝试此方法，它将起作用。