如何在 Scala 中验证数字字符？
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How do I validate a Numeric character in scala?
提问by yash
my application takes in a string like this (-110,23,-111.9543633) I need to validate/retrieve inside scala script that the string whether it is Numeric or not?
我的应用程序接受这样的字符串 (-110,23,-111.9543633) 我需要在 Scala 脚本中验证/检索该字符串是否为数字？
采纳答案by yash
I tried this and its working fine:
我试过这个并且它工作正常：
val s = "-1112.12" s.isNumeric && (s.contains('.')==false)
val s = "-1112.12" s.isNumeric && (s.contains('.')==false)
回答by elm
Consider scala.util.Tryfor catching possible exceptions in converting a string onto a numerical value, as follows,
考虑scala.util.Try在将字符串转换为数值时捕获可能的异常，如下所示，
Try("123".toDouble).isSuccess
Boolean = true
Try("a123".toDouble).isSuccess
Boolean = false
As of ease of use, consider this implicit,
至于易用性，请考虑这个隐含的，
implicit class OpsNum(val str: String) extends AnyVal {
def isNumeric() = scala.util.Try(str.toDouble).isSuccess
}
Hence
因此
"-123.7".isNumeric
Boolean = true
"-123e7".isNumeric
Boolean = true
"--123e7".isNumeric
Boolean = false
回答by Sascha Kolberg
Assuming you want not only to convert by using Try(x.toInt)or Try(x.toFloat)but actually want a validatorthat takes a Stringand return trueiff the passed Stringcan be converted to a Awith implicit evidence: Numeric[A].
假设您不仅想要通过 using 进行转换，Try(x.toInt)或者Try(x.toFloat)实际上想要一个接受 a并返回的验证器，如果传递的可以转换为with 。StringtrueStringAimplicit evidence: Numeric[A]
Then I would say: Afaik, noit is not possible. Especially, since Numericis not sealed. That is you can create your own implementations of Numeric
然后我会说：Afaik，不，这是不可能的。尤其是，因为Numeric不是密封的。也就是说，您可以创建自己的实现Numeric
Primitive types like Int, Long, Float, Doublecan be easily extracted with a regular expression.
Int, Long, Float, Double可以使用正则表达式轻松提取诸如此类的原始类型。
