javax.ws.rs.WebApplicationException:com.sun.jersey.api.MessageException:Java REST Web服务中的JSON支持

Question

好吧,之前可能已经提出过这个问题了,但是在所有网站上我都看到了"如何做到这一点"的解释告诉我,我做得完全正确.

我知道我不是,因为我的localhost tomcat上出现500服务器错误,我的服务器日志中出现以下错误:

javax.ws.rs.WebApplicationException: com.sun.jersey.api.MessageException: A message body writer for Java class com.myapp.domain.Location, and Java type class com.myapp.domain.Location, and MIME media type application/json was not found

因此,我要做的是使用Jersey(Java)开发RESTful Web服务.一切都很顺利,除了我想要返回JSON的事实.我无法找到与这些人不同的东西:

• 如何在Jersey Rest中以JSON形式发送响应
• http://www.jasonwhaley.com/blog/2011/01/18/multiple-content-types-in-jax-rs/
• https://github.com/jasonray/jersey-starterkit/wiki/Serializing-a-POJO-to-xml-or-json-using-JAXB
• https://github.com/jasonray/jersey-starterkit/wiki/Serializing-a-POJO-to-json-using-built-in-jersey-support
• http://www.vogella.com/tutorials/REST/article.html

我的POJO(位置)看起来像这样:

package com.myapp.domain;

import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement()
public class Location {
    private int id;
    private double longtitude;
    private double latitude;

    public Location() {
        new Location(-1, -1, -1);
    }

    public Location(double longtitude, double latitude) {
        new Location(-1, longtitude, latitude);
    }

    public Location(int id, double longtitude, double latitude) {
        this.id = id;
        this.longtitude = longtitude;
        this.latitude = latitude;
    }

    public void setID(int id) {
        this.id = id;
    }

    public void setLongtitude(double longtitude) {
        this.longtitude = longtitude;
    }

    public void setLatitude(double latitude) {
        this.latitude = latitude;
    }

    public int getID() {
        return this.id;
    }

    public double getLongtitude() {
        return this.longtitude;
    }

    public double getLatitude() {
        return this.latitude;
    }
}

我的资源看起来像这样:

package com.myapp.MyAPP;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;

import com.myapp.domain.Location;

@Path("Locations")
public class LocationInfo {
    @GET
    @Path("/get/{id}")
    @Produces(MediaType.APPLICATION_JSON)
    public Location getLocation(@PathParam("id") int id) {
        Location loc = new Location(3, 4.007391, 51.00237);
        return loc;
    }
}

这是我的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" 
    xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" 
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
        http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
        id="WebApp_ID" 
        version="2.5">  
    <display-name>MyAPP</display-name> 
    <servlet> 
            <servlet-name>MyAPP REST Service</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>com.myapp.MyAPP</param-value>
            </init-param>
            <init-param>
                <param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
                <param-value>true</param-value>
            </init-param>
            <load-on-startup>1</load-on-startup> 
    </servlet> 
    <servlet-mapping> 
            <servlet-name>MyAPP REST Service</servlet-name> 
            <url-pattern>/rest/*</url-pattern> 
    </servlet-mapping>
</web-app>

我有这些库包括:asm-3.1.jar,jersey-client-1.17.1.jar,jersey-core-1.17.1.jar,jersey-json-1.17.1.jar,jersey-server-1.17.1.jar,jersey-servlet-1.17.jar,jsr11-api-1.1.1.jar

看到我没看到的人得到了啤酒.或者至少是我永恒的感激之情,因为我一直在看这个太久了,我仍然看不到它.

Answer 1

好的，所以我发现 orid 有正确的答案：我只需要添加一些额外的库！

我可能忽略了它，或者大多数教程可能会假设您立即添加了使用 jersey 下载的所有库...

所以这解决了问题：添加

• jackson-core-asl-1.9.2.jar
• jackson-jaxrs-1.9.2.jar
• jackson-mapper-asl-1.9.2.jar
• jackson-xc-1.9.2.jar

再次感谢 orid，你拯救了我的周末。