我的程序有很多不同的功能,其中一个是命令“load”。 一旦用户输入“load”输入,他就可以加载到txt文件中...... 问题是,我的命令不仅仅是“加载”字本身,例如“load numbers.txt”或“load data.txt”
现在我想打开这些位于我的电脑上的文本文件,但我需要在命令前面没有“加载”的文件名。如何从整个输入行中仅获取名称?
def ProgramSelector() {
var endProgram = false
while (!endProgram) {
val userSelection = scala.io.StdIn.readLine("There is no transfer data available yet, please use the 'load' command to initialize the application!\nEnter your command or type 'help' for more information:")
if (userSelection == "help")
println("some help text here")
else if (userSelection == "load")
//else if (userSelection == "3")
//exerciseThree()
//else if (userSelection == "4")
//exerciseFour()
//else if (userSelection == "5")
//exerciseFive()
//else if (userSelection == "6")
//exerciseSix()
//else if (userSelection == "7")
//exerciseSeven()
//else if (userSelection == "8")
//exerciseEight()
else if (userSelection == "exit")
endProgram = true
else
println("Invalid command!")
所以我的函数是ProgramSelector,如果输入是加载的话我只会创建一个if语句...
答案 0 :(得分:1)
我试图让这更通用一点。
为了说明这有什么用处,我还创建了另一个命令,你可以称之为“add 1 2”,它将打印添加两个整数的总和。
如果您认真制作CLI交互式应用程序,我建议您查看如何在sbt之上创建自己的交互式shell的here。
val loadCommand = """load (.*)""".r
val helpCommand = """help.*""".r
val exitCommand = """exit.*""".r
val addCommand = """add\s+(\d+)\s+(\d+)""".r
val PromptMsg = "There is no transfer data available yet, please use the 'load' command to initialize the application!\nEnter your command or type 'help' for more information: "
def programSelector() {
var endProgram = false
val fileKeeper = new scala.collection.mutable.HashSet[String]()
while (!endProgram) {
val userSelection = scala.io.StdIn.readLine(PromptMsg)
userSelection match {
case loadCommand(file) =>
println(s"Adding file $file")
fileKeeper add file
println(s"Files so far: $fileKeeper")
case helpCommand() =>
println("some help text here")
case exitCommand() =>
endProgram = true
case addCommand(a,b) =>
val sum = a.toInt + b.toInt
println(s"Sum=$sum")
case _ =>
println("Invalid command!")
}
}
}
programSelector()