在Java中使用访问和密钥调用AWS API网关

我正在尝试使用Java调用具有访问权限和密钥的AWSAPI网关端点。使用邮递员可以很好地工作。下面是我的邮递员输入

GET
Url - https://example.com/testing
Authorization:
Type - AWS Signature
AccessKey - xxxxxxxxxxxx
SecretKey - yyyyyyyyyyyy
AWS Region - us-east-1
Service Name - execute-api

在使用Java调用时，由于出现禁止的错误消息而失败。不知道问题出在哪里。我知道这与授权有关。我看过AWS文件，但不清楚。谢谢你的帮助

这是代码

import java.util.Date;
import java.text.SimpleDateFormat;
import java.util.List;
import java.net.*;
import javax.crypto.Mac;
import javax.crypto.spec.SecretKeySpec;
import java.io.*;

public class InvokeApi {

    static byte[] HmacSHA256(String data, byte[] key) throws Exception {
        String algorithm = "HmacSHA256";
        Mac mac = Mac.getInstance(algorithm);
        mac.init(new SecretKeySpec(key, algorithm));
        return mac.doFinal(data.getBytes("UTF-8"));
    }

    static byte[] getSignatureKey(String key, String dateStamp, String regionName, String serviceName)
            throws Exception {
        byte[] kSecret = ("AWS4" + key).getBytes("UTF-8");
        byte[] kDate = HmacSHA256(dateStamp, kSecret);
        byte[] kRegion = HmacSHA256(regionName, kDate);
        byte[] kService = HmacSHA256(serviceName, kRegion);
        byte[] kSigning = HmacSHA256("aws4_request", kService);
        return kSigning;

    }

    public static void main(String[] args) {

        String newurl = "https://example.com/testing";
        String key = "yyyyyyyyyyyy";
        String dateStamp = "2020-11-04 12:28:12+00:00";
        String regionName = "us-east-1";
        String serviceName = "execute-api";
        String awsAccessKeyID = "xxxxxxxxxxxx";
        String completeResponse = "";

        URL url;
        try {
            url = new URL(newurl);
            HttpURLConnection conn = (HttpURLConnection) url.openConnection();
            conn.setRequestMethod("GET");
            conn.setRequestProperty("Accept", "*/*");
            byte getSignatureKey[] = getSignatureKey(key, dateStamp, regionName, serviceName);

            conn.setRequestProperty("Authorization", "AWS " + awsAccessKeyID + ":" + getSignatureKey);
            if (conn.getResponseCode() == 200) {
                InputStreamReader inputStream = new InputStreamReader(conn.getInputStream());
                BufferedReader br = new BufferedReader(inputStream);
                String output;
                while ((output = br.readLine()) != null) {
                    completeResponse += output;
                }
                System.out.println("completeResponse :" + completeResponse);
                br.close();
                inputStream.close();
                if (conn != null) {
                    conn.disconnect();
                }
            } else {
                System.out.println("Eror code : " + conn.getResponseMessage());
            }
        } catch (MalformedURLException e) {
            e.printStackTrace();
            System.out.println("Exception 1 : " + e);
        } catch (IOException e) {
            e.printStackTrace();
            System.out.println("Exception 2 : " + e);
        } catch (Exception e) {
            e.printStackTrace();
            System.out.println("Exception 3 : " + e);
        }
    }
}

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为什么不直接使用aws sdk呢？我已经尝试过了，但是没有成功。如果有一个好的例子，我将不胜感激。aws文档非常好。在源代码中放置秘密也是一种非常糟糕的做法。您应该使用aws cli和安装配置文件。谢谢，让我看一下文档。是的，我使用个人资料，这里只是为了清楚。