Snowflake Scala Connector（Snekark）

Question

我是Scala的新手，并尝试使用雪花雪板Scala库在Scala中创建雪花连接器。

这是我提供了正确凭据的简单代码

package com.abc.commons.rest.snowflake
import com.snowflake.snowpark._
import com.snowflake.snowpark.functions._

object ScalaConnector {
  def main(args: Array[String]): Unit = {
    // Replace the <placeholders> below.
    val configs = Map (
      "URL" -> "https://xxx.snowflakecomputing.com:443",
      "USER" -> "xxx",
      "PASSWORD" -> "xxx",
      "ROLE" -> "xxx",
      "WAREHOUSE" -> "xxx",
      "DB" -> "xxx",
      "SCHEMA" -> "xxx"
    )
    val session = Session.builder.configs(configs).create
    session.sql("show tables").show()
    session.close();
  }
}

，然后在Intellij中运行上述代码时，我的错误：

Exception in thread "main" java.lang.ExceptionInInitializerError
    at com.abc.commons.rest.snowflake.ScalaConnector$.main(ScalaConnector.scala:17)
    at com.abc.commons.rest.snowflake.ScalaConnector.main(ScalaConnector.scala)
Caused by: java.lang.NullPointerException
    at com.snowflake.snowpark.Session$.getActiveSession(Session.scala:1204)
    at com.snowflake.snowpark.SnowparkClientException.<init>(SnowparkClientException.scala:15)
    at com.snowflake.snowpark.internal.ErrorMessage$.createException(ErrorMessage.scala:380)
    at com.snowflake.snowpark.internal.ErrorMessage$.MISC_SCALA_VERSION_NOT_SUPPORTED(ErrorMessage.scala:340)
    at com.snowflake.snowpark.internal.Utils$.checkScalaVersionCompatibility(Utils.scala:243)
    at com.snowflake.snowpark.internal.Utils$.checkScalaVersionCompatibility(Utils.scala:233)
    at com.snowflake.snowpark.Session$.<init>(Session.scala:1129)
    at com.snowflake.snowpark.Session$.<clinit>(Session.scala)
    ... 2 more

liblary Dippedencies 我在 build.sbuild.sbt < /strong>

"com.snowflake" % "snowpark" % "1.4.0"

有人可以在我的代码中指出问题吗？

Answer 1

代码没有任何问题，您只是使用不支持的Scala版本。从您的stacktrace中：

misc_scala_version_not_supported

我们只支持scala 2.12，如解释。