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检测负载的regex模式。*模式

scala regex

0

Chaitanya · 技术社区 · 8 年前

我要验证所有具有“payload.*”的字符串,即该字符串必须以“payload”开头,后跟句点(.)和最少1个字符。

示例:

Input1 :- payload.Hello     Output1 :-> Valid
Input2 :- Hipayload.Hello     Output1 :-> InValid
Input3 :- payload.H     Output1 :-> Valid
Input4 :- payload.     Output1 :-> InValid

3 回复 | 直到 8 年前

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James Whiteley 8 年前

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

2

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AGR 8 年前

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\.[^.]+$

3

0

RAGHHURAAMM 8 年前

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)