如何在spring应用程序中编写REST api?我收到了这个错误:
此请求标识的资源只能生成 根据请求具有不可接受的特征的回复 "接受"头。
我尝试了很多。我的代码如下所示:
MVC-调度-servlet.xml中
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:mvc="http://www.springframework.org/schema/mvc" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.1.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.1.xsd
http://www.springframework.org/schema/mvc
http://www.springframework.org/schema/mvc/spring-mvc-3.1.xsd">
<context:component-scan base-package="com.mkyong.common.controller" />
<mvc:annotation-driven />
<bean
class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix">
<value>/WEB-INF/pages/</value>
</property>
<property name="suffix">
<value>.jsp</value>
</property>
</bean>
</beans>
的web.xml
<web-app id="WebApp_ID" version="2.4"
xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee
http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>Spring Web MVC Application</display-name>
<servlet>
<servlet-name>mvc-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/movie/*</url-pattern>
</servlet-mapping>
</web-app>
控制器类文件
package com.mkyong.common.controller;
import java.util.ArrayList;
import java.util.List;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.bind.annotation.ResponseBody;
import com.mkyong.common.model.UserBean;
@Controller
@RequestMapping("/movie")
public class MovieController {
@RequestMapping(value="/response", method = RequestMethod.GET)
public @ResponseBody List<UserBean> getMovie() {
UserBean bean = new UserBean();
List<UserBean> list = new ArrayList<UserBean>();
bean.setId("xx");
bean.setFirstname("xxx");
bean.setLastname("xxx");
bean.setSal("xxxx");
list.add(bean);
return list;
}
}
UserBean类
package com.mkyong.common.model;
public class UserBean {
private String id;
private String firstname;
private String lastname;
private String sal;
public String getId() {
return id;
}
public void setId(String id) {
this.id = id;
}
public String getFirstname() {
return firstname;
}
public void setFirstname(String firstname) {
this.firstname = firstname;
}
public String getLastname() {
return lastname;
}
public void setLastname(String lastname) {
this.lastname = lastname;
}
public String getSal() {
return sal;
}
public void setSal(String sal) {
this.sal = sal;
}
}
答案 0 :(得分:0)
Spring无法将对象转换为JSON。确保您的类路径中有jackson-databind
。如果您使用的是Maven,请将其添加到pom.xml
:
<dependency>
<groupId>com.fasterxml.jackson.core</groupId>
<artifactId>jackson-databind</artifactId>
<version>2.4.3</version>
</dependency>
答案 1 :(得分:0)
您可以通过在控制器上使用@RequestMapping的“consume”和“produce”属性来指定请求/响应的映射类型。例如,如果您在json中发送请求,并且您还在json中收到响应,则应使用
@RequestMapping(value="/response", method = RequestMethod.GET, consumes = "application/json", produces = "application/json" )
检查您的请求:
您的错误是指“接受”标题,因此我认为问题是您需要指定“生产”属性
答案 2 :(得分:0)
最后,这位内容谈判代表工作了:
<mvc:annotation-driven content-negotiation-manager="contentNegotiationManager" />