使用spring发送和接收json数据

Question

我正在使用GOOGLE CHROME的POST MAN客户端发送articleName和articleId AS HEADER应用程序/ json.What我需要在我的控制器和库以及我的spring servlet中进行更改。 xml？我的控制器如下。

公共类ArticleController {

@Autowired 
    private ArticleService articleService; 

    Article article = new Article(); 
    Long articleId = article.getArticleId(); 

    @RequestMapping(value = "/save", method = RequestMethod.POST) 

    public Article saveArticle(@ModelAttribute Article article, 
      BindingResult bindingresult) { 

     int a = articleService.addArticle(article); 
     if (a == 1) { 
      return new ModelAndView("success"); 
     } else { 
      return new ModelAndView("error"); 
     } 
    } 


My Spring servlet is: 
<?xml version="1.0" encoding="UTF-8"?> 
<beans xmlns="http://www.springframework.org/schema/beans" 
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:context="http://www.springframework.org/schema/context" 
    xmlns:tx="http://www.springframework.org/schema/tx" 
    xmlns:mvc="http://www.springframework.org/schema/mvc" 
    xsi:schemaLocation=" 
http://www.springframework.org/schema/beans 
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd 
http://www.springframework.org/schema/context 
http://www.springframework.org/schema/context/spring-context-3.0.xsd 
http://www.springframework.org/schema/tx 
http://www.springframework.org/schema/tx/spring-tx-3.0.xsd 
http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd"> 

    <context:property-placeholder location="classpath:jdbc.properties" /> 
    <context:component-scan base-package="net.roseindia" /> 

    <tx:annotation-driven transaction-manager="hibernateTransactionManager"/> 
    <mvc:annotation-driven /> 


    <bean id="jspViewResolver" 
     class="org.springframework.web.servlet.view.InternalResourceViewResolver"> 
     <property name="viewClass" 
      value="org.springframework.web.servlet.view.JstlView" /> 
     <property name="prefix" value="/WEB-INF/view/" /> 
     <property name="suffix" value=".jsp" /> 
    </bean> 






    <bean id="dataSource" 
     class="org.springframework.jdbc.datasource.DriverManagerDataSource"> 
     <property name="driverClassName" value="${database.driver}" /> 
     <property name="url" value="${database.url}" /> 
     <property name="username" value="${database.user}" /> 
     <property name="password" value="${database.password}" /> 
    </bean> 

    <bean id="sessionFactory" 
     class="org.springframework.orm.hibernate3.annotation.AnnotationSessionFactoryBean"> 
     <property name="dataSource" ref="dataSource" /> 
     <property name="annotatedClasses"> 
      <list> 
       <value>net.roseindia.model.Article</value> 
      </list> 
     </property> 
     <property name="hibernateProperties"> 
      <props> 
       <prop key="hibernate.dialect">${hibernate.dialect}</prop> 
       <prop key="hibernate.show_sql">${hibernate.show_sql}</prop>    
      </props> 
     </property> 
    </bean> 

    <bean id="hibernateTransactionManager" 
     class="org.springframework.orm.hibernate3.HibernateTransactionManager"> 
     <property name="sessionFactory" ref="sessionFactory" /> 
    </bean> 
</beans> 

PLZ帮我....预先感谢。

Answer 1

我假设你想要的是从Spring轻松返回JSON。

要做到这一点，你需要杰克逊依赖：

public @ResponseBody Article saveArticle(@ModelAttribute Article article, 
     BindingResult bindingresult) { 
.... 
} 

这种方法将返回JSONified：

<dependency> 
     <groupId>org.codehaus.jackson</groupId> 
     <artifactId>jackson-mapper-asl</artifactId> 
     <version>1.7.1</version> 
    </dependency> 

当你拥有它， 你可以@ResponseBody注释就像这个注解你的方法条款对象的回应。