regex - Scala，正则表达式匹配忽略不必要的单词

Question

我的程序是:

val pattern = "[*]prefix_([a-zA-Z]*)_[*]".r

val outputFieldMod = "TRASHprefix_target_TRASH"

var tar =
  outputFieldMod match {
    case pattern(target) => target
  }

println(tar)

基本上，我尝试获取“目标”并忽略“TRASH”(我使用了*)。但它有一些错误，我不知道为什么..

Answer 1

简单直接的标准库函数(未锚定)

使用非锚定

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解决方案一

在模式上使用unanchored来匹配字符串内部，忽略垃圾

val pattern = "prefix_([a-zA-Z]*)_".r.unanchored

unanchored 将仅匹配忽略所有垃圾(所有其他单词)的模式

val result = str match {
 case pattern(value) => value
 case _ => ""
}

示例

Scala REPL

scala> val pattern = """foo\((.*)\)""".r.unanchored
pattern: scala.util.matching.UnanchoredRegex = foo\((.*)\)

scala> val str = "blahblahfoo(bar)blahblah"
str: String = blahblahfoo(bar)blahblah

scala> str match { case pattern(value) => value ; case _ => "no match" }
res3: String = bar

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解决方案二

用.*从两侧填充图案。 .* 匹配除换行符之外的任何字符。

val pattern = ".*prefix_([a-zA-Z]*)_.*".r

val result = str match {
   case pattern(value) => value
   case _ => ""
}

示例

Scala REPL

scala> val pattern = """.*foo\((.*)\).*""".r
pattern: scala.util.matching.Regex = .*foo\((.*)\).*

scala> val str = "blahblahfoo(bar)blahblah"
str: String = blahblahfoo(bar)blahblah

scala> str match { case pattern(value) => value ; case _ => "no match" }
res4: String = bar