在 Java Play 框架中实施 AWS Secret Manager

Question

我正在尝试在java play框架中实现aws秘密管理器。我按照本文https://www.geekyhacker.com/2020/05/09/getting-database-credentials-from-aws-secrets-manager-in-spring-boot/中的步骤首先在springboot中实现它工作完美。

现在，我在 play 框架中遵循相同的步骤。

第1步：我在build.sbt中添加了以下依赖项

libraryDependencies ++= Seq(
  "com.amazonaws" % "aws-java-sdk" % "1.12.122",
  "com.amazonaws.secretsmanager" % "aws-secretsmanager-jdbc" % "1.0.6",
  "com.fasterxml.jackson.module" %% "jackson-module-scala" % "2.12.2"
)

其他相关依赖项包括：

libraryDependencies ++= Seq(
  "mysql" % "mysql-connector-java" % "8.0.21"
)
libraryDependencies ++= Seq(
  javaJpa,
  "org.hibernate" % "hibernate-core" % "5.4.21.Final", // it is JPA implementation
  "javax.validation" % "validation-api" % "2.0.1.Final" // added this because an issue was coming while running
)

步骤2：更新conf文件

旧配置（运行良好）

play.db {
  default = unclassified
}
base_db_url = "jdbc:mysql://{my_rds_endpoint}/"
db_annotations= "?createDatabaseIfNotExist=true"
base_db_user_name= "{rds_user_name}"
base_db_password= "{rds_password}"

db {
  unclassified.driver = com.mysql.cj.jdbc.Driver//com.mysql.cj.jdbc.Driver
  unclassified.jndiName = UnclassifiedDS
  unclassified.url= ${base_db_url}"unclassified_"${serverEnvironment}${db_annotations}
  unclassified.username=${base_db_user_name}
  unclassified.password=${base_db_password}
}

新配置：

play.db {
      default = unclassified
    }
    base_db_url = "jdbc-secretsmanager:mysql://{my-rds-endpoint}/"
    db_annotations= "?createDatabaseIfNotExist=true"
    base_db_user_name= "{secret-name in secret manager}"
db {    
    unclassified.driver=com.amazonaws.secretsmanager.sql.AWSSecretsManagerMySQLDriver
    unclassified.jndiName = UnclassifiedDS
    unclassified.url= ${base_db_url}"unclassified_"${serverEnvironment}${db_annotations}
    unclassified.username=${base_db_user_name}
}

但出现错误：配置错误[无法初始化数据库] 原因：java.lang.IllegalArgumentException：需要dataSource或dataSourceClassName或jdbcUrl。

有人能帮我一下吗？