Question

我有一个包含字符串的列表。例如：10001_20180101010101_SOURCE_Text

我想对所有字符串进行if语句，例如第一个字符串的ID大于第二个字符串的ID，然后是print("anything")，依此类推。

我尝试过这样的模式匹配：

val pattern = raw"([0-9]+)_(\d{4})(\d{2})(\d{2})(\d{2})(\d{2})(\d{2})_(SOURCE|DEL)_(.*)".r
"some string" match {
  case pattern(id, year, month, day, hour, min, sec, typee, name: String) => 
    println(s"$id $year $month $day $hour $min $sec $typee $name")
  case _ => println("No match")
}

我可以以某种方式访问ID，年，月...吗？

Answer 1

我有一个包含字符串的列表。例如：“ 10001_20180101010101_SOURCE_Text”我想为所有字符串创建if语句...如果第一个字符串的ID大于第二个字符串的ID，则进行print（“ anything”），依此类推。

像这样吗？

case class Record(id: Int, year: Int, month: Int)

val pattern = raw"([0-9]+)_(\d{4})(\d{2})(\d{2})(\d{2})(\d{2})(\d{2})_(TEXT|text|Text)_(.*)".r

def parse(record): Option[Record] =
  "some string" match {
    case pattern(id, year, month, day, hour, min, sec, typee, name: String) => 
      Some(Record(id, year, month))
      // println(s"$id $year $month $day $hour $min $sec $typee $name")
    case _ => None
      // println("No match")

val inputRecords: List[String] = ... // from input

val records: List[Record] = inputRecords.flatMap(parse)

if (records.get(0).id > records.get(1).id)
   print("anything")

模式匹配Scala，访问值

1 个答案: