Question

我目前正在使用Eclipse和AWS Toolkit for Eclipse。我的项目已经正常工作，它正在完成它的工作，即连接到RDS实例并将JSON对象返回到API网关调用。

我刚刚获得了新的要求，我们将使用服务SecretsManager自动轮换RDS配置，例如用户，密码等。

问题是当我尝试导入GetSecretValueResponse等类时，我得到The import com.amazonaws.services.secretsmanager cannot be resolved。当我浏览文档和SDK时，存在GetSecretValueRequest而不是GetSecretValueResponse，所以我无法理解我应该做什么，也没有找到类似于示例的内容我可以学习。

以下代码是我正在尝试实现的内容，由亚马逊自己提供（在Secrets Manager页面中有一个按钮，您可以单击以查看它将如何与Java一起使用，在这种情况下），并呈现没有任何修改，因为正如我所说，我不知道如何导入几个类：

// Use this code snippet in your app.
public static void getSecret() {
String secretName = "secretName";
String endpoint = "secretEndpoint";
String region = "region";

AwsClientBuilder.EndpointConfiguration config = new AwsClientBuilder.EndpointConfiguration(endpoint, region);
AWSSecretsManagerClientBuilder clientBuilder = AWSSecretsManagerClientBuilder.standard();
clientBuilder.setEndpointConfiguration(config);
AWSSecretsManager client = clientBuilder.build();

String secret;
ByteBuffer binarySecretData;
GetSecretValueRequest getSecretValueRequest = GetSecretValueRequest.builder()
        .withSecretId(secretName)
        .build();
GetSecretValueResponse getSecretValueResponse = null;
try {
    getSecretValueResponse = client.getSecretValue(getSecretValueRequest);

} catch(ResourceNotFoundException e) {
    System.out.println("The requested secret " + secretName + " was not found");
} catch (InvalidRequestException e) {
    System.out.println("The request was invalid due to: " + e.getMessage());
} catch (InvalidParameterException e) {
    System.out.println("The request had invalid params: " + e.getMessage());
}

if(getSecretValueResponse == null) {
    return;
}

// Decrypted secret using the associated KMS CMK
// Depending on whether the secret was a string or binary, one of these fields will be populated
if(getSecretValueResponse.getSecretString() != null) {
    secret = getSecretValueResponse.getSecretString();
}
else {
    binarySecretData = getSecretValueResponse.getSecretBinary();
}

// Your code goes here. 
}

Answer 1

我遇到了同样的问题，AWS页面上的代码并不是开箱即用的。您要找的班级是GetSecretValueResult 这是最新的java文档

https://docs.aws.amazon.com/AWSJavaSDK/latest/javadoc/com/amazonaws/services/secretsmanager/model/GetSecretValueResult.html

这是一件可行的作品：

public void printRdsSecret() throws IOException {
    String secretName = "mySecretName";

    System.out.println("Requesting secret...");
    AWSSecretsManager client = AWSSecretsManagerClientBuilder.standard().build();

    GetSecretValueRequest getSecretValueRequest = new GetSecretValueRequest().withSecretId(secretName);

    GetSecretValueResult getSecretValueResult = client.getSecretValue(getSecretValueRequest);

    System.out.println("secret retrieved ");
    final String secretBinaryString = getSecretValueResult.getSecretString();
    final ObjectMapper objectMapper = new ObjectMapper();

    final HashMap<String, String> secretMap = objectMapper.readValue(secretBinaryString, HashMap.class);

    String url = String.format("jdbc:postgresql://%s:%s/dbName", secretMap.get("host"), secretMap.get("port"));
    System.out.println("Secret url = "+url);
    System.out.println("Secret username = "+secretMap.get("username"));
    System.out.println("Secret password = "+secretMap.get("password"));
 }

使用版本aws-java-sdk-secretsmanager

的1.11.337对此进行了测试

Answer 2

只需将lib添加到pom： https://mvnrepository.com/artifact/com.amazonaws/aws-java-sdk-secretsmanager

Answer 3

我认为主要问题是缺少对 AWS SDK v2 的依赖。

在此处添加使用 AWS SDK v2 的代码段。万一有人在寻找这个。

package com.may.util;

import software.amazon.awssdk.regions.Region;
import software.amazon.awssdk.services.secretsmanager.SecretsManagerClient;
import software.amazon.awssdk.services.secretsmanager.model.DecryptionFailureException;
import software.amazon.awssdk.services.secretsmanager.model.GetSecretValueRequest;
import software.amazon.awssdk.services.secretsmanager.model.GetSecretValueResponse;
import software.amazon.awssdk.services.secretsmanager.model.InternalServiceErrorException;
import software.amazon.awssdk.services.secretsmanager.model.InvalidParameterException;
import software.amazon.awssdk.services.secretsmanager.model.InvalidRequestException;
import software.amazon.awssdk.services.secretsmanager.model.ResourceNotFoundException;

public class SecretsManagerUtil {

    public static String obtainSecret() {
        String secretName = "db_secret_name";
        String region = "us-east-1";

        SecretsManagerClient client = SecretsManagerClient.builder().region(Region.of(region)).build();
        GetSecretValueResponse response = null;

        try {
            response = client.getSecretValue(GetSecretValueRequest.builder().secretId(secretName).build());
        } catch (DecryptionFailureException e) {
            // Secrets Manager can't decrypt the protected secret text using the provided KMS key.
            // Deal with the exception here, and/or rethrow at your discretion.
            throw e;
        } catch (InternalServiceErrorException e) {
            // An error occurred on the server side.
            // Deal with the exception here, and/or rethrow at your discretion.
            throw e;
        } catch (InvalidParameterException e) {
            // You provided an invalid value for a parameter.
            // Deal with the exception here, and/or rethrow at your discretion.
            throw e;
        } catch (InvalidRequestException e) {
            // You provided a parameter value that is not valid for the current state of the resource.
            // Deal with the exception here, and/or rethrow at your discretion.
            throw e;
        } catch (ResourceNotFoundException e) {
            // We can't find the resource that you asked for.
            // Deal with the exception here, and/or rethrow at your discretion.
            throw e;
        }

        return response.secretString();
    }
}

反序列化并打印机密：

public class SecretPrinter {

private static final Logger logger = LoggerFactory.getLogger(SecretPrinter.class);

public void printSecret() {
    String json = SecretsManagerUtil.obtainSecret(); // secret in json format

    RdsSecret secret;
    try {
        secret = new ObjectMapper().disable(FAIL_ON_UNKNOWN_PROPERTIES).readValue(json, RdsSecret.class);
    } catch (IOException e) {
        logger.error("Couldn't parse secret obtained from AWS Secrets Manager!");
        throw new RuntimeException(e);
    }

    System.out.println("username: " + secret.getUsername());
    System.out.println("password: " + secret.getPassword());
}

static class RdsSecret {
    private String username;
    private String password;

    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }
}

}

行家：

<dependencyManagement>
  <dependencies>
    <dependency>
      <groupId>software.amazon.awssdk</groupId>
      <artifactId>bom</artifactId>
      <version>2.6.3</version>
      <type>pom</type>
      <scope>import</scope>
    </dependency>
  </dependencies>
</dependencyManagement>

<dependency>
  <groupId>software.amazon.awssdk</groupId>
  <artifactId>secretsmanager</artifactId>
</dependency>

等级：

implementation platform('software.amazon.awssdk:bom:2.6.3')
implementation 'software.amazon.awssdk:secretsmanager'

Answer 4

我建议使用aws secret manger jdbc包装器访问RDS（https://github.com/aws/aws-secretsmanager-jdbc）。在传递给RDS客户端之前，您无需处理获取秘密，解码并使用文本中的密码来回移动。

只需将秘密ID传递给RDS客户端，然后jdbc包装程序将处理其余部分。

Answer 5

aws-secretsmanager-jdbc 可用于通过AWS Secrete Manager访问AWS RDS。 https://github.com/aws/aws-secretsmanager-jdbc

下面是我的application.properties的内容

spring.datasource.url=jdbc-secretsmanager:mysql://dev-xxxx-database.cluster-xxxxxxxxx.ap-southeast-1.rds.amazonaws.com:3306/dev_xxxxxx

spring.datasource.username=/secret/application
spring.datasource.driver-class-name=com.amazonaws.secretsmanager.sql.AWSSecretsManagerMySQLDriver

spring.jpa.database-platform = org.hibernate.dialect.MySQL5Dialect

以下是AWS Secret Manager中的秘密。

使用此方法，您无需手动获取用户名和密码并创建数据源。

Answer 6

您缺少“ aws-java-sdk-secretsmanager”依赖性，必须将其添加到pom.xml中，然后导入到Java类中。

来自Maven：

    <dependency>
        <groupId>com.amazonaws</groupId>
        <artifactId>aws-java-sdk-secretsmanager</artifactId>
        <version>1.11.355 </version>
    </dependency>

AWS Reference

如果您不使用Maven，则您有add AWS SDK to your existing project.

使用AWS Secrets Manager

6 个答案: