Question

我正在尝试使用此链接创建Web sersive http://www.programming-free.com/2014/03/spring-mvc-40-restful-web-service-json.html 我按照所有步骤操作，因为我需要静态数据，所以我不想连接数据库。但是我收到错误

我按照步骤

1）新增 - ＆gt;动态网络项目

2）添加url中提供的所有库

共享记录-1.1.1.jar

    jackson-annotations-2.0.0.jar
    jackson-core-2.0.0.jar
    jackson-databind-2.0.0.jar
    jstl-1.2.jar
    mysql-connector-java-3.1.12.jar
    spring-aop-4.0.0.RELEASE.jar
    spring-aspects-4.0.0.RELEASE.jar
    spring-beans-4.0.0.RELEASE.jar
    spring-context-4.0.0.RELEASE.jar
    spring-core-4.0.0.RELEASE.jar
    spring-expression-4.0.0.RELEASE.jar
    spring-framework-bom-4.0.0.RELEASE.jar
    spring-web-4.0.0.RELEASE.jar
    spring-webmvc-4.0.0.RELEASE.jar

的web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
  <display-name>SpringServiceJsonSample</display-name>
  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
  </welcome-file-list>
  <servlet>
    <servlet-name>rest</servlet-name>
    <servlet-class>
        org.springframework.web.servlet.DispatcherServlet
    </servlet-class>
    <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>rest</servlet-name>
    <url-pattern>/</url-pattern>
  </servlet-mapping>
</web-app>

其余-servlet.xml中

<?xml version="1.0" encoding="UTF-8"?>
<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"
    xmlns:p="http://www.springframework.org/schema/p"
    xsi:schemaLocation="
        http://www.springframework.org/schema/beans     
        http://www.springframework.org/schema/beans/spring-beans-4.0.xsd
        http://www.springframework.org/schema/context
        http://www.springframework.org/schema/context/spring-context-4.0.xsd
        http://www.springframework.org/schema/mvc
        http://www.springframework.org/schema/mvc/spring-mvc-4.0.xsd">
    <context:component-scan base-package="com.programmingfree.springservice.controller" />
    <mvc:annotation-driven />
  </beans>

controller.js

package com.programmingfree.springservice.controller;

import java.util.HashMap;

import org.springframework.web.bind.annotation.PathVariable;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.bind.annotation.RestController;

@RestController
@RequestMapping("/service/user/")
public class SpringServiceController {

    HashMap<String, String> userService=new HashMap<String, String>();

    @RequestMapping(value = "/test", method = RequestMethod.GET,headers="Accept=application/json")
    public HashMap getUser(@PathVariable int id) {
        userService.put("name", "hello");
        return userService;
    }




}

当我点击服务器时的 http://localhost:8080/dddddd/service/user/test 它显示客户端发送的请求在语法上是不正确的（）。

Answer 1

你的mpaping需要一个PathVariable id，但是你的映射模式没有定义它

将其添加到请求映射或从参数中删除它，映射将成功。以下是已删除PathVariable的示例

 @RequestMapping(value = "/test", method =  equestMethod.GET,headers="Accept=application/json")
public HashMap getUser() {
 }

如何删除此错误“客户端发送的请求在语法上是不正确的（）。”

1 个答案: