spark-submit 程序中scala和python的构造

scala

linux 环境
spark 2.4.2
scala 2.12.8 （spark-scala版本通过运行spark-shell查看）

简单运行

目录结构

find .
.
./hw.scala

创建hw.scala文件

object Hi{
        def main(args: Array[String]) = println("Hello world")
}

在当前目录下直接运行
sbt
然后sbt>中输入 run
即得到结果

构造项目

目录结构

$ find .
.
./simple.sbt
./src
./src/main
./src/main/scala
./src/main/scala/SimpleApp.scala

其中SimapleApp.scala的内容：
需要修改logFile的文件目录

/* SimpleApp.scala */
import org.apache.spark.SparkContext
import org.apache.spark.SparkContext._
import org.apache.spark.SparkConf

object SimpleApp {
  def main(args: Array[String]) {
    val logFile = "YOUR_SPARK_HOME/README.md"
    val conf = new SparkConf().setAppName("Simple Application")
    val sc = new SparkContext(conf)
    val logData = sc.textFile(logFile, 2).cache()
    val numAs = logData.filter(line => line.contains("a")).count()
    val numBs = logData.filter(line => line.contains("b")).count()
    println(s"Lines with a: $numAs, Lines with b: $numBs")
    sc.stop()
  }
}

simple.sbt是配置文件：
注意要有空行

name := "Simple Project"

version := "1.0"

scalaVersion := "2.12.8"

libraryDependencies += "org.apache.spark" %% "spark-core" % "2.4.2"

然后再主目录下
sbt package

最后运行
spark-submit --class “SimpleApp” target/scala-2.12/simple-project_2.12-1.0.jar

Ref:

1. https://www.jianshu.com/p/454cb5318372
2. https://www.cnblogs.com/codingexperience/p/5372617.html