在尝试JSON使用我的响应时Spring 3.x,我得到406 error"此请求标识的资源只能生成具有根据request"accept"标题()接受的特性不可接受的响应."
这是我的环境
* Spring 3.2.0.RELEASE
* included jackson-mapper-asl-1.7.9.jar, jackson-core-asl-1.7.9.jar
* Tomcat 6.x
* mvc:annotation-driven in Spring configuration XML file
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我的控制器:
@RequestMapping("/contest")
public class ContestController {
@RequestMapping(value="{name}", headers="Accept=*/*", method = RequestMethod.GET)
public @ResponseBody Contest getContestInJSON(@PathVariable String name) {
Contest contest = new Contest();
contest.setName(name);
contest.setStaffName(new String("contestitem1"));
return contest;
}
}
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我的Spring配置文件
<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-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd
http://www.springframework.org/schema/mvc
http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd">
<context:component-scan base-package="com.contestframework.controllers" />
<bean class="org.springframework.web.servlet.view.ContentNegotiatingViewResolver">
<property name="mediaTypes">
<map> …Run Code Online (Sandbox Code Playgroud)