Question

我正在编写Scala代码来搜索特定文件，如果reg-ex匹配则在该文件中返回该行。为简单起见，我编写了一个简单的正则表达式。我的代码是

import scala.io.Source
import scala.util.matching
object regex {
 println("Welcome to the Scala worksheet")

   val patter=""".*(BEGINNING).*""".r




   val dirName="/home/ugan/shaileshsir/20aug/temp.dwuKM/asup.txt"

 //var n=new 

 //for (filen <- nt){
try {
 for (line <- Source.fromFile(dirName).getLines()) {
   //println(line)
  line match
   {
   case patter(group) => group
    case _ => ""

   }
   println(line)
  }
} catch {
  case ex: Exception => println("Bummer, an exception happened.")
 }

虽然println(line)工作正常并打印文件但我无法找到与正则表达式匹配的行。任何建议都会有所帮助

Answer 1

我的建议是：

GUID

如果你的正则表达式只是一个Source.fromFile(dirName).getLines().filter(_ matches ".*(BEGINNING).*")，你也可以考虑这个：

.*{Something}.*

Answer 2

import java.io._
object firstprogram {
    def main(args: Array[String]) {
        val regex = ".*(BEGINNING).*".r
        val filename = "ScalaFile.txt"  
        var fileSource = scala.io.Source.fromFile(filename).getLines  
        print(fileSource+"\n")
        for(x<-fileSource){
            var fileyo=x.replaceAll(".*(BEGINNING).*","x")
            val fileObject = new PrintWriter(new FileOutputStream(new File("ScalaFiletest1.txt"),true))   //true For appending
            val printWriter = new PrintWriter(fileObject)
            printWriter.write(fileyo+"\n")  // Writing to the new file  
            printWriter.close()
        }
    }
}

如何在Scala中查找文件中的特定模式

2 个答案: