我正在使用GOOGLE CHROME的POST MAN CLIENT发送articleName和articleId AS HEADER application / json。我需要在我的控制器和库以及我的spring servlet.xml中更改哪些内容?我的控制器如下。
public class 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帮助我....提前致谢。
答案 0 :(得分:0)
我认为你想要的是从Spring轻松返回JSON。
要做到这一点,你需要杰克逊依赖:
<dependency>
<groupId>org.codehaus.jackson</groupId>
<artifactId>jackson-mapper-asl</artifactId>
<version>1.7.1</version>
</dependency>
当你拥有它时,
您可以使用@ResponseBody
注释对此方法进行注释,如下所示:
public @ResponseBody Article saveArticle(@ModelAttribute Article article,
BindingResult bindingresult) {
....
}
这样的方法将返回JSONified Article对象作为响应。