使用Ajax将@RequestBody中的多个变量传递给Spring MVC控制器 [英] Passing multiple variables in @RequestBody to a Spring MVC controller using Ajax

问题描述

是否需要包裹背衬物体？我想这样做：

Is it necessary to wrap in a backing object? I want to do this:

@RequestMapping(value = "/Test", method = RequestMethod.POST)
@ResponseBody
public boolean getTest(@RequestBody String str1, @RequestBody String str2) {}

并使用这样的JSON：

And use a JSON like this:

{
    "str1": "test one",
    "str2": "two test"
}

但我必须使用：

@RequestMapping(value = "/Test", method = RequestMethod.POST)
@ResponseBody
public boolean getTest(@RequestBody Holder holder) {}

然后使用这个JSON：

And then use this JSON:

{
    "holder": {
        "str1": "test one",
        "str2": "two test"
    }
}

这是正确的吗？我的另一个选择是将 RequestMethod 更改为 GET 并使用 @RequestParam 在查询字符串中或使用 @PathVariable 与 RequestMethod 。

Is that correct? My other option would be to change the RequestMethod to GET and use @RequestParam in query string or use @PathVariable with either RequestMethod.

推荐答案

你是对的，@ RequestBody带注释的参数应该保存整个请求并绑定到一个对象，所以你基本上必须选择。

You are correct, @RequestBody annotated parameter is expected to hold the entire body of the request and bind to one object, so you essentially will have to go with your options.

如果你绝对想要你的方法，你可以做一个自定义的实现：

If you absolutely want your approach, there is a custom implementation that you can do though:

说这是你的json：

{
    "str1": "test one",
    "str2": "two test"
}

和你一起想把它绑定到这两个参数：

and you want to bind it to the two params here:

@RequestMapping(value = "/Test", method = RequestMethod.POST)
public boolean getTest(String str1, String str2)

首先定义一个自定义注释，比如说 @JsonArg ，其中包含JSON路径，例如您想要的信息路径：

First define a custom annotation, say @JsonArg, with the JSON path like path to the information that you want:

public boolean getTest(@JsonArg("/str1") String str1, @JsonArg("/str2") String str2)

现在写一个Custom HandlerMethodArgumentResolver 使用 JsonPath 定义以上解决实际参数：

Now write a Custom HandlerMethodArgumentResolver which uses the JsonPath defined above to resolve the actual argument:

import java.io.IOException;

import javax.servlet.http.HttpServletRequest;

import org.apache.commons.io.IOUtils;
import org.springframework.core.MethodParameter;
import org.springframework.http.server.ServletServerHttpRequest;
import org.springframework.web.bind.support.WebDataBinderFactory;
import org.springframework.web.context.request.NativeWebRequest;
import org.springframework.web.method.support.HandlerMethodArgumentResolver;
import org.springframework.web.method.support.ModelAndViewContainer;

import com.jayway.jsonpath.JsonPath;

public class JsonPathArgumentResolver implements HandlerMethodArgumentResolver{

    private static final String JSONBODYATTRIBUTE = "JSON_REQUEST_BODY";
    @Override
    public boolean supportsParameter(MethodParameter parameter) {
        return parameter.hasParameterAnnotation(JsonArg.class);
    }

    @Override
    public Object resolveArgument(MethodParameter parameter, ModelAndViewContainer mavContainer, NativeWebRequest webRequest, WebDataBinderFactory binderFactory) throws Exception {
        String body = getRequestBody(webRequest);
        String val = JsonPath.read(body, parameter.getMethodAnnotation(JsonArg.class).value());
        return val;
    }

    private String getRequestBody(NativeWebRequest webRequest){
        HttpServletRequest servletRequest = webRequest.getNativeRequest(HttpServletRequest.class);
        String jsonBody = (String) servletRequest.getAttribute(JSONBODYATTRIBUTE);
        if (jsonBody==null){
            try {
                String body = IOUtils.toString(servletRequest.getInputStream());
                servletRequest.setAttribute(JSONBODYATTRIBUTE, body);
                return body;
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return "";

    }
}

现在只需在Spring MVC中注册。有点涉及，但这应该干净利落。

Now just register this with Spring MVC. A bit involved, but this should work cleanly.

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