检测负载的regex模式。*模式

你可以用 x.matches(y) (何处) x 和 y 都是字符串)以与scala中的正则表达式模式匹配(或使用“原始”字符串):

scala> val regex: String = "^payload\\..+"
regex: String = ^payload\..+

scala> val regexAltRaw: String = raw"^payload\..+"
regexAltRaw : String = ^payload\..+

scala> val regexAltTripleQuotes: String = """^payload\..+"""
regexAltTripleQuotes : String = ^payload\..+

scala> "".matches(regex)
res0: Boolean = false

scala> "payload.Hello".matches(regex)
res1: Boolean = true

scala> "Hipayload.Hello".matches(regex)
res2: Boolean = false

scala> "payload.H".matches(regex)
res3: Boolean = true

scala> "payload.".matches(regex)
res4: Boolean = false

为了解释这种模式:

• ^payload -以“有效载荷”开头
• \\. -”字面意思(如果不使用“原始”字符串,scala要求您使用双反斜杠来转义,而不是像在普通regex中那样使用单斜杠)
• .+ -任何角色,一次或多次

为了避免验证输入有效载荷。您可以使用愚蠢的正则表达式:

^payload\.[^.]+$

输入数组:

val ar = Array("payload.Hello","Hipayload.Hello","payload.H","payload.")

正则表达式字符串:

 val p = """^payload\..{1,}""" 

在scala repl中:

scala> val ar = Array("payload.Hello","Hipayload.Hello","payload.H","payload.")
ar: Array[String] = Array(payload.Hello, Hipayload.Hello, payload.H, payload.)

scala> val p = """^payload\..{1,}"""
p: String = ^payload\..{1,}

测试:

scala> ar.map(x=>if(x.matches(p))"Valid" else "InValid")
res3: Array[String] = Array(Valid, InValid, Valid, InValid)