HttpServletRequest 无法获取 xhr 请求参数

Question

我正在使用 Spring 和 javascript。使用 XhrHttpRequest 对象调用 @Controller。

我可以使用 Chrome Inspector 查看我的参数（JSON 字符串），但是当我调用 request.getParamter("id") 时返回 null。

Calling part with js
function ajax(url, data, callback, method){
    //data is {"id":"system", "password" : "1234"}
    var httpRequest;
    var afterAction = function(){
        if(!httpRequest) {
            console.error('can not find httpRequest variable');
            return;
        }

        if (httpRequest.readyState === XMLHttpRequest.DONE) {
            if (httpRequest.status === 200) {
                var responseData = httpRequest.responseText;
                //alert(JSON.stringify(responseData));
                console.log('Result of API call >>>', responseData);
                if(typeof callback == 'function') {
                    callback(JSON.parse(responseData));
                }
            } else {
                alert('There was a problem with the request.');
            }
        }
    }

    //=========== LOGIC ============
    if (window.XMLHttpRequest) { // Mozilla, Safari, IE7+ ...
        httpRequest = new XMLHttpRequest();
    } else if (window.ActiveXObject) { // IE 6 and older
        httpRequest = new ActiveXObject('Microsoft.XMLHTTP');
    }

    if(!method) method = 'POST';
    data = (!!data) ? JSON.stringify(data) : '';

    httpRequest.onreadystatechange = afterAction;
    httpRequest.open(method.toUpperCase(), url, true);
    httpRequest.setRequestHeader("Content-type", "application/json; charset=utf-8");
    //httpRequest.setRequestHeader("Content-length", data.length);
    //httpRequest.setRequestHeader("Connection", "close");
    httpRequest.send(data);
}

使用 Spring @Controller 接收零件

@Controller
@RequestMapping(value={"member"}, produces={"application/json"})
@ResponseStatus(value = HttpStatus.OK)
public class MemberController {

    /**
     * @param request
     * @param resp
     * @return
     * @throws Exception
     */
    @RequestMapping(value={"/login"})
    public @ResponseBody String login(HttpServletRequest request, HttpServletResponse resp) throws Exception {
        System.out.println("Login request");
        String id = String.valueOf(request.getParameter("id"));                 //returns null
        String password = String.valueOf(request.getParameter("password"));     //returns null

        Map<String, String> result = new HashMap<String, String>();
        result.put("result", "S");
        result.put("message", "login success");
        ObjectMapper mapper = new ObjectMapper();
        return mapper.writeValueAsString(result);
    }
}

我不知道为什么参数变为空。谢谢。

Answer 1

这是杰克逊图书馆的另一种方式。我喜欢 @javaguy 的方式，但使用 Jackson 库可能更灵活，所以添加这篇文章。

准备工作

• WebMvcConfigurerAdapter 实现类或与之等效的东西
• SpringFramework 环境。在我的例子中，Eclipse Dynamic Web Project 与 Maven 和 spring-webmvc 在 pom.xml 中。

我做了什么...

• 实现WebMvcConfigurerAdapter
• 覆盖方法configureMessageConverters
• 将Controller参数从HttpServeletRequest request, HttpServletResponse resp更改为@RequestBody Map<?, ?>。（您可以更改通用值。进一步处理无关紧要）

下面是代码。

ServerConfig.java（我阻止其他选项以关注当前问题）

@Configuration
@EnableWebMvc
@ComponentScan(basePackages={ ... }
, excludeFilters=@ComponentScan.Filter(Configuration.class))
//Filter 걸 때 Configuration.class 를 수동으로 등록해줘야 되는데 나은 방법 찾아보기 
public class ServerConfig extends WebMvcConfigurerAdapter {

    ...

    @Override
    public void configureMessageConverters(List<HttpMessageConverter<?>> converters) {
        converters.add(new MappingJackson2HttpMessageConverter());
        super.configureMessageConverters(converters);
    }

};

和控制器（更改参数类型）

@Controller
@RequestMapping(value={"member"}, produces={"application/json"})
@ResponseStatus(value = HttpStatus.OK)
public class MemberApiController {

    ...

    @RequestMapping(value={"/login"})
    public @ResponseBody String login(@RequestBody Map<?, ?> jsonBody) throws JsonProcessingException {
        System.out.println("Login request" + jsonBody.toString());
        Map<String, String> result = new HashMap<String, String>();
        String rs = "S";
        String message = "SUCCESS";

        System.out.println("ID >>> ", String.valueOf(jsonBody.get("id")));
        System.out.println("PW >>> ", String.valueOf(jsonBody.get("password")));

        result.put("result", rs);
        result.put("result_msg", message);
        mapper = new ObjectMapper();
        return mapper.writeValueAsString(result);
    }
};

但是，据我所知@RequestBody没有HttpSession，如果你尝试将数据存储到HttpSession，你需要另一个参数HttpServletRequest。

我希望这可以帮助与我有同样问题的其他人：D

Answer 2

您需要按照以下步骤在您的控制器中接受Json：

(1) 定义 UserLogin bean 来保存 Json

public class UserLogin  {
    private String id;
    private String password;

    //Add getters and setters
}

(2) 更改您的控制器以接受 Json 并接收 UserLogin bean

@Controller
@RequestMapping(value={"member"}, produces={"application/json"})
@ResponseStatus(value = HttpStatus.OK)
public class MemberController {

    @RequestMapping(value={"/login"}, method = RequestMethod.POST, consumes = MediaType.APPLICATION_JSON_VALUE)
    public @ResponseBody String login(UserLogin userLogin) throws Exception {
        System.out.println("Login request");
        String id = userLogin.getId();            
        String password = userLogin.getPassword();

        Map<String, String> result = new HashMap<String, String>();
        result.put("result", "S");
        result.put("message", "login success");
        ObjectMapper mapper = new ObjectMapper();
        return mapper.writeValueAsString(result);
    }
}