java Jersey JSON 序列化

Jersey JSON serialization

提问by KGolbang

I set up an eclipse WebApp project and placed Jersey and Hymanson JARs in the WEB-INF/lib directory. I want to use JSON serialization but didn't manage to fix this error:

我设置了一个 Eclipse WebApp 项目并将 Jersey 和 Hymanson JAR 放在 WEB-INF/lib 目录中。我想使用 JSON 序列化，但未能修复此错误：

MessageBodyWriter not found for media type=application/json, type=class com.rest.Greeting, genericType=class com.rest.Greeting. I already googled much but all the solutions are updated or don't solve my issue. Here is the structure of my project:

The WEB-INF/libfolder contains following JARs:

在WEB-INF / lib目录文件夹包含以下JAR：

aopalliance-repackaged-2.3.0-b10.jar
asm-debug-all-5.0.2.jar
cglib-2.2.2.jar
datafactory-0.8.jar
hk2-api-2.3.0-b10.jar
hk2-locator-2.3.0-b10.jar
hk2-utils-2.3.0-b10.jar
Hymanson-annotations-2.4.0.jar
Hymanson-core-2.4.0.jar
Hymanson-databind-2.4.0.jar
Hymanson-jaxrs-base-2.4.0.jar
Hymanson-jaxrs-json-provider-2.4.0.jar
Hymanson-module-jaxb-annotations-2.4.0.jar
javassist-3.18.1-GA.jar
javax.annotation-api-1.2.jar
javax.inject-2.3.0-b10.jar
javax.servlet-api-3.0.1.jar
javax.ws.rs-api-2.0.1.jar
jaxb-api-2.2.7.jar
jersey-client.jar
jersey-common.jar
jersey-container-servlet-core.jar
jersey-container-servlet.jar
jersey-guava-2.13.jar
jersey-server.jar
org.osgi.core-4.2.0.jar
osgi-resource-locator-1.0.1.jar
persistence-api-1.0.jar
validation-api-1.1.0.Final.jar

web.xml:

网页.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>Rest Test</display-name>
  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
  </welcome-file-list>
  <servlet>
    <servlet-name>Greeting</servlet-name>
    <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
    <init-param>
        <param-name>jersey.config.server.provider.packages</param-name>
        <param-value>com.rest</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>Greeting</servlet-name>
    <url-pattern>/greet/*</url-pattern>
  </servlet-mapping>
</web-app>

Greeting.java:

问候.java:

@XmlRootElement
@Path("/greeting")
public class Greeting {

    private String greeting;

    public Greeting() {
        this.setGreeting("hello");
    }

    @GET
    @Produces(MediaType.TEXT_XML)   
    public Greeting sayHello() {
        return new Greeting();
    }

    public String getGreeting() {
        return greeting;
    }

    public void setGreeting(String greeting) {
        this.greeting = greeting;
    }

}

The WebApp project is run in an embedded Tomcat v7 server (view "Servers" in eclipse). The request:

WebApp 项目在嵌入式 Tomcat v7 服务器中运行（在 Eclipse 中查看“服务器”）。 请求：

http://localhost:8080/com.rest/greet/greeting

As long as I use MediaType.TEXT_XMLthere is no error and the content looks like this:

只要我使用MediaType.TEXT_XML就没有错误，内容如下所示：

<greeting>
<greeting>hello</greeting>
</greeting>

When I change MediaType to "MediaType.APPLICATION_JSON"then the following exception (as described above) is thrown. I already noticed that it depends on the annotation "@XmlRootElement" but there is not something like "JsonRootElement" and the Hymanson/genson APIs claim that it works out of the box (due to the auto registration mechanism). Do you have any idea how to fix it?

当我将 MediaType 更改为“MediaType.APPLICATION_JSON”时，会引发以下异常（如上所述）。我已经注意到它取决于注释“@XmlRootElement”，但没有像“JsonRootElement”这样的东西，Hymanson/genson API 声称它开箱即用（由于自动注册机制）。你知道如何解决它吗？

Every help is very very appreciated! thank you golbie

每一个帮助都非常非常感谢！谢谢你

回答by Lukasz Wiktor

All you need is to register HymansonJsonProvider. There are many ways to achieve that:

您所需要的只是注册HymansonJsonProvider。有很多方法可以实现：

1. Register the HymansonJsonProvider explicitly in web.xml:

1. 在 web.xml 中显式注册 HymansonJsonProvider：

<init-param>
    <param-name>jersey.config.server.provider.classnames</param-name>
    <param-value>com.fasterxml.Hymanson.jaxrs.json.HymansonJsonProvider</param-value>
</init-param>

or

或者

1. Register your class extending javax.ws.rs.core.Applicationin the web.xml

2. 注册您javax.ws.rs.core.Application在 web.xml 中扩展的类

<init-param>
    <param-name>javax.ws.rs.Application</param-name>
    <param-value>com.rest.MyApplication</param-value>
</init-param>

and then do all the configuration in the application class:

然后在应用程序类中进行所有配置：

package com.rest;

import org.glassfish.jersey.server.ResourceConfig;
import com.fasterxml.Hymanson.jaxrs.json.HymansonJsonProvider;

public class MyApplication extends ResourceConfig {

    public MyApplication() {
        packages("com.rest");
        register(HymansonJsonProvider.class);
    }

ResourceConfigis a subclass of javax.ws.rs.Applicationand gives you some helper methods that makes the registration easy.

ResourceConfig是 的子类，javax.ws.rs.Application并为您提供一些使注册变得容易的辅助方法。

or

或者

1. Use automatic registration. Just add dependency to jersey-media-json-Hymanson

3. 使用自动注册。只需添加依赖jersey-media-json-Hymanson

<dependency>
    <groupId>org.glassfish.jersey.media</groupId>
    <artifactId>jersey-media-json-Hymanson</artifactId>
    <version>2.13</version>
</dependency>

But be careful. It will register more than you need:

不过要小心。它将注册比您需要的更多：

• HymansonJaxbJsonProvider,
• JsonParseExceptionMapper,
• JsonMappingExceptionMapper

• HymansonJaxbJsonProvider,
• JsonParseExceptionMapper，
• JsonMappingExceptionMapper

Take a look at the source codeto see what it does.

查看源代码以了解它的作用。

回答by mendieta

Have you tried adding this to your web.xml?

您是否尝试将其添加到您的 web.xml 中？

<init-param>
    <param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
    <param-value>true</param-value>
</init-param>

I believe this tutorial shows what you are trying yo do http://examples.javacodegeeks.com/enterprise-java/rest/jersey/json-example-with-jersey-Hymanson/

我相信本教程显示了您正在尝试做的事情 http://examples.javacodegeeks.com/enterprise-java/rest/jersey/json-example-with-jersey-Hymanson/

Another one from mkyong http://www.mkyong.com/webservices/jax-rs/json-example-with-jersey-Hymanson/

另一个来自 mkyong http://www.mkyong.com/webservices/jax-rs/json-example-with-jersey-Hymanson/