在Scala中创建Java对象 [英] Creating a Java Object in Scala

问题描述

我有一个Java类"Listings".我在Java MapReduce作业中使用它，如下所示:

I have a Java class "Listings". I use this in my Java MapReduce job as below:

public void map(Object key, Text value, Context context) throws IOException, InterruptedException {
Listings le = new Listings(value.toString());
...
}

我想在Spark上运行相同的作业.因此，我现在正在Scala中编写此代码.我导入了Java类:

I want to run the same job on Spark. So, I am writing this in Scala now. I imported the Java class:

import src.main.java.lists.Listings

我想在Scala中创建一个Listings对象.我正在这样做:

I want to create a Listings object in Scala. I am doing this:

val file_le = sc.textFile("file// Path to file")
Listings lists = new Listings(file_le)

我得到一个错误:

值列表不是对象src.main.java.lists.Listings的成员

value lists is not a member of object src.main.java.lists.Listings

正确的方法是什么?

推荐答案

基于您的发言，我认为您可能会忘记Scala语法和Java语法之间的区别.

Based on what you've said, I think you may be forgetting the differences between Scala syntax and Java syntax.

尝试一下:

val lists: Listings = new Listings(SomeString)

请注意，在Scala中指定类型是完全可选的.另外，如果您要更改列表的值，请使用var.

Please note that specifying the type in Scala is completely optional. Also, use a var if you're going to be changing the value of lists.

您拥有它的方式，Scala试图通过其调用不带'.'的对象的方法/访问值的能力来解释它，因此您实际上是在告诉Scala:

The way you have it, Scala is trying to interpret it by its ability to call methods/access values of an object without the '.', so you're actually telling Scala this:

Listings.lists = new Listings(SomeString)

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