snakeyaml和spark导致无法构造对象

Question

在给定snakeyaml版本1.17的scala shell中,以下代码执行正常

import org.yaml.snakeyaml.Yaml
import org.yaml.snakeyaml.constructor.Constructor
import scala.collection.mutable.ListBuffer
import scala.beans.BeanProperty

class EmailAccount {
  @scala.beans.BeanProperty var accountName: String = null

  override def toString: String = {
    return s"acct ($accountName)"
  }
}

val text = """accountName: Ymail Account"""

val yaml = new Yaml(new Constructor(classOf[EmailAccount]))
val e = yaml.load(text).asInstanceOf[EmailAccount]
println(e)

但是当在spark(在这种情况下为2.0.0)中运行时,产生的错误是:

org.yaml.snakeyaml.constructor.ConstructorException: Can't construct a java object for tag:yaml.org,2002:EmailAccount; exception=java.lang.NoSuchMethodException: EmailAccount.<init>()
 in 'string', line 1, column 1:
    accountName: Ymail Account
    ^

  at org.yaml.snakeyaml.constructor.Constructor$ConstructYamlObject.construct(Constructor.java:350)
  at org.yaml.snakeyaml.constructor.BaseConstructor.constructObject(BaseConstructor.java:182)
  at org.yaml.snakeyaml.constructor.BaseConstructor.constructDocument(BaseConstructor.java:141)
  at org.yaml.snakeyaml.constructor.BaseConstructor.getSingleData(BaseConstructor.java:127)
  at org.yaml.snakeyaml.Yaml.loadFromReader(Yaml.java:450)
  at org.yaml.snakeyaml.Yaml.load(Yaml.java:369)
  ... 48 elided
Caused by: org.yaml.snakeyaml.error.YAMLException: java.lang.NoSuchMethodException: EmailAccount.<init>()
  at org.yaml.snakeyaml.constructor.Constructor$ConstructMapping.createEmptyJavaBean(Constructor.java:220)
  at org.yaml.snakeyaml.constructor.Constructor$ConstructMapping.construct(Constructor.java:190)
  at org.yaml.snakeyaml.constructor.Constructor$ConstructYamlObject.construct(Constructor.java:346)
  ... 53 more
Caused by: java.lang.NoSuchMethodException: EmailAccount.<init>()
  at java.lang.Class.getConstructor0(Class.java:2810)
  at java.lang.Class.getDeclaredConstructor(Class.java:2053)
  at org.yaml.snakeyaml.constructor.Constructor$ConstructMapping.createEmptyJavaBean(Constructor.java:216)
  ... 55 more

我用scora shell启动了

scala -classpath "/home/placey/snakeyaml-1.17.jar"

我推出了火花壳

/home/placey/Downloads/spark-2.0.0-bin-hadoop2.7/bin/spark-shell --master local --jars /home/placey/snakeyaml-1.17.jar

Answer 1

解

创建一个自包含的应用程序并使用spark-submit而不是使用它来运行它spark-shell.

我在这里为你创建了一个最小的项目.您需要做的就是将两个文件(build.sbt和Main.scala)放在某个目录中,然后运行:

sbt package

为了创建一个JAR.JAR将在target/scala-2.11/sparksnakeyamltest_2.11-1.0.jar或类似的位置.如果你还没有使用它,你可以从这里获得SBT.最后,您可以运行该项目:

/home/placey/Downloads/spark-2.0.0-bin-hadoop2.7/bin/spark-submit --class "Main" --master local --jars /home/placey/snakeyaml-1.17.jar target/scala-2.11/sparksnakeyamltest_2.11-1.0.jar

输出应该是:

[many lines of Spark's log)]
acct (Ymail Account)
[more lines of Spark's log)]

说明

Spark的shell(REPL)通过向$iw构造函数添加参数来转换您在其中定义的所有类.我在这里解释过.SnakeYAML期望类似JavaBean类的零参数构造函数,但没有一个,所以它失败了.

你可以自己试试:

scala> class Foo() {}
defined class Foo

scala> classOf[Foo].getConstructors()
res0: Array[java.lang.reflect.Constructor[_]] = Array(public Foo($iw))

scala> classOf[Foo].getConstructors()(0).getParameterCount
res1: Int = 1

如您所见,Spark通过添加类型的参数来转换构造函数$iw.

替代方案

定义自己的 Constructor

如果你真的需要让它在shell中工作,你可以定义你自己的类实现org.yaml.snakeyaml.constructor.BaseConstructor并确保$iw传递给构造函数,但这是很多工作(我Constructor出于安全原因实际上在Scala中编写了我自己的类,不久前,所以我对此有一些经验).

您还可以定义一个自定义Constructor硬编码来实例化特定类(EmailAccount在您的情况下),类似于DiceConstructor SnakeYAML文档中所示.这样更容易,但需要为要支持的每个类编写代码.

例:

case class EmailAccount(accountName: String)

class EmailAccountConstructor extends org.yaml.snakeyaml.constructor.Constructor {

  val emailAccountTag = new org.yaml.snakeyaml.nodes.Tag("!emailAccount")
  this.rootTag = emailAccountTag
  this.yamlConstructors.put(emailAccountTag, new ConstructEmailAccount)

  private class ConstructEmailAccount extends org.yaml.snakeyaml.constructor.AbstractConstruct {
    def construct(node: org.yaml.snakeyaml.nodes.Node): Object = {
      // TODO: This is fine for quick prototyping in a REPL, but in a real
      //       application you should probably add type checks.
      val mnode = node.asInstanceOf[org.yaml.snakeyaml.nodes.MappingNode]
      val mapping = constructMapping(mnode)
      val name = mapping.get("accountName").asInstanceOf[String]
      new EmailAccount(name)
    }
  }

}

您可以将其另存为文件并使用在REPL中加载:load filename.scala.

此解决方案的额外优势在于它可以直接创建不可变的案例类实例.不幸的是,Scala REPL似乎存在导入问题,因此我使用了完全限定名称.

不要使用JavaBeans

您也可以将YAML文档解析为简单的Java映射:

scala> val yaml2 = new Yaml()
yaml2: org.yaml.snakeyaml.Yaml = Yaml:1141996301

scala> val e2 = yaml2.load(text)
e2: Object = {accountName=Ymail Account}

scala> val map = e2.asInstanceOf[java.util.Map[String, Any]]
map: java.util.Map[String,Any] = {accountName=Ymail Account}

scala> map.get("accountName")
res4: Any = Ymail Account

这样SnakeYAML就不需要使用反射了.

但是,由于您使用的是Scala,我建议您尝试使用 MoultingYAML,它是SnakeYAML的Scala包装器.它将YAML文档解析为简单的Java类型,然后将它们映射到Scala类型(甚至是您自己的类型EmailAccount).