如何在 Scala 中解析导入的名称?(火花/齐柏林飞艇) [英] How is an imported name resolved in Scala? (Spark / Zeppelin)

问题描述

我在 Zeppelin 中使用 Spark 解释器在一个段落中运行了一个脚本.它有一个导入，导入的名称可以从全局命名空间和函数解析，但不能从类中的方法解析.

I have a script running in a paragraph with the Spark interpreter in Zeppelin. It has an import and the name imported can be resolved from the global namespace and also from a function, but not from a method inside a class.

这在我计算机上安装的 Scala (2.12) 上运行良好，但在 Zeppelin (Scala 2.11) 中不起作用.

This runs well on my computer's installation of Scala (2.12) but it doesn't work in Zeppelin (Scala 2.11).

import java.util.Calendar

def myFun: String = {
    // this works
    return Calendar.getInstance.toString
}

class MyClass {
    def myFun(): String = {
      // this doesn't
      return Calendar.getInstance.toString

      // this works
      return java.util.Calendar.getInstance.toString
    }
}

错误信息如下:

import java.util.Calendar
myFun: String
<console>:15: error: not found: value Calendar
                 return Calendar.getInstance.toString

我错过了什么?

推荐答案

在 0.8.0 中，Zeppelin 引入了一个新的 SparkInterpreter，我想是因为全局导入不起作用，导入必须在包装器内进行限定.

In 0.8.0 Zeppelin has introduced a new SparkInterpreter I suppose due to which the global imports don't work and the imports have to be scoped within a wrapper.

作为一种解决方法，可以将属性 zeppelin.spark.useNew 设置为值false".这会禁用 sparks 新解释器.

As a workaround, the property zeppelin.spark.useNew can be set to the value "false". This disables the sparks new interpreter.

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