
正文
sprngmvc+restFul 请求报错:404
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服务端代码
control类
@RequestMapping(value="getUser",method = RequestMethod.POST)
@ResponseBody
public User getUser(HttpServletRequest request){
String name = request.getParameter("name");
String age = request.getParameter("age");
User user = new User();
user.setPassWord("123123");
user.setUserName("asdasd");
return user;
}
spring核心控制器
<!--配置Springmvc核心控制器 -->
<servlet>
<servlet-name>springmvc</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:springMVC.xml</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>springmvc</servlet-name>
<url-pattern>*.action</url-pattern>
</servlet-mapping>
客户端代码
import com.alibaba.fastjson.JSONObject;
import org.apache.http.HttpResponse;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.StringEntity;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.util.EntityUtils;
import org.junit.Test; import java.io.IOException; public class HttpclientTest {
@Test
public void httpReqUrl()
throws ClientProtocolException, IOException {
String url = "http://localhost:8080/ssm/user/getUser";
JSONObject json = new JSONObject();
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost method = new HttpPost(url);
// 接收参数json列表
JSONObject jsonParam = new JSONObject(); StringEntity entity = new StringEntity(jsonParam.toString(), "utf-8");//解决中文乱码问题
entity.setContentEncoding("UTF-8");
entity.setContentType("application/json");
method.setEntity(entity); HttpResponse result = httpClient.execute(method);
// 请求结束,返回结果
String resData = EntityUtils.toString(result.getEntity());
JSONObject resJson = json.parseObject(resData);
System.out.printf(":::"+resJson.toString());
}
}
调用报错:http.404
报错原因:服务端配置的核心控制器,只拦截以 “.action”结尾的请求
解决方案:将服务端核心控制器配置“<url-pattern>*.action</url-pattern>”改为“<url-pattern>/</url-pattern>”








