java - 在 Spring MVC 中使用 HttpServletRequest 之前替换/准备原始 POST 数据

Question

我有基于 api 的客户端-服务器应用程序，php+yii 作为后端。我的目标是将后端部分的端口从 php 移植到 java(与来自客户端的旧查询兼容)。

我选择了 Spring MVC 框架，但在解密原始发布数据时遇到了问题。问题是旧客户端使用自定义加密算法，因此 HttpServletRequest.getParameterMap() 不会返回有效映射。

所以，我的问题是:在这种情况下，在对纯业务逻辑使用 request.getParameterMap() 方法之前替换/准备原始发布数据(解密)的最佳解决方案是什么？

Controller 代码是这样的:

@RestController
@RequestMapping(value = "/", method = RequestMethod.POST)
public class ApiController {

    @RequestMapping
    @ResponseBody
    public String process(HttpServletRequest request) {
        ApiQueryResponse queryResponse = new ApiQueryResponse();

        try {
            // request = APIEncryptor.decrypt(request);         // 1
            // now working with request.getParameterMap()       // 2
            // ...
            queryResponse.setText(APIEncryptor.crypt(...))
        } catch (Exception e) {
            queryResponse.setError(e.getMessage());
        }

        return queryResponse.asText();
    }

    ...

}

APIEncryptor 代码是这样的:

public class ApiEncryptor {
    public static ApiQueryData encrypt(Java Object) {
        // custom encryption algorithm...
        return Raw data string;
    }

    // input post data: 103klg20fsl2g2fsldkfj20f9isjf
    // output post data: a=1&b=2&c=3&d=some_text

    public static ApiQueryData decrypt(Raw post data...) {
        // decrypt algorithm...
        return request; // HttpServletRequest object with valid post data
        // after this action, .getParameter() will produce valid data
    }
}

注意:我知道存在什么方法，例如request.getInputStream()，但我现在不需要它。我需要知道在使用 HttpServletRequest 参数 getter 之前是否有任何方法可以准备发布数据？或者(无论如何)我需要一次从请求中获取发布数据并做我想做的事(例如通过 getReader())？

Answer 1

The best way to handle your decryption logic is inside a http filter (uses spring MVC) as below. This filter decrypts the requests immediately after receiving the request to server and before reaching the controller method.

import java.io.IOException;
import javax.servlet.FilterChain;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import org.springframework.web.filter.OncePerRequestFilter;

public class DecryptionFilter extends OncePerRequestFilter {

    @Override
    protected void doFilterInternal(HttpServletRequest request, HttpServletResponse response, FilterChain filterChain)
            throws ServletException, IOException {

        request = ApiEncryptor.decrypt(request);

        filterChain.doFilter(request, response);
    }

}