Scala 子字符串函数
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/44443212/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Scala subString function
提问by John
Hi I am looking for a solution it will return a substring from string for the given indexes.For avoiding index bound exception currently using if and else check.Is there a better approach(functional).
嗨，我正在寻找一种解决方案，它将为给定的索引从字符串返回一个子字符串。为了避免当前使用 if 和 else 检查的索引绑定异常。是否有更好的方法（功能）。
def subStringEn(input:String,start:Int,end:Int)={
// multiple if check for avoiding index out of bound exception
input.substring(start,end)
}
回答by Tzach Zohar
Not sure what you want the function to do in case of index out of bound, but slicemight fit your needs:
不确定在索引超出范围的情况下您希望该函数做什么，但slice可能符合您的需求：
input.slice(start, end)
Some examples:
一些例子：
scala> "hello".slice(1, 2)
res6: String = e
scala> "hello".slice(1, 30)
res7: String = ello
scala> "hello".slice(7, 8)
res8: String = ""
scala> "hello".slice(0, 5)
res9: String = hello
回答by prayagupd
Tryis one way of doing it. The other way is applying substring only if length is greater than end using Option[String].
Try是一种方法。另一种方法是仅当长度大于 end using 时才应用子字符串Option[String]。
invalid end index
无效的结束索引
scala> val start = 1
start: Int = 1
scala> val end = 1000
end: Int = 1000
scala> Option("urayagppd").filter(_.length > end).map(_.substring(start, end))
res9: Option[String] = None
valid end index
有效的结束索引
scala> val end = 6
end: Int = 6
scala> Option("urayagppd").filter(_.length > end).map(_.substring(start, end))
res10: Option[String] = Some(rayag)
Also, you can combine filterand mapto .collectas below,
此外，您还可以结合filter并map以.collect如下，
scala> Option("urayagppd").collect { case x if x.length > end => x.substring(start, end) }
res14: Option[String] = Some(rayag)
scala> val end = 1000
end: Int = 1000
scala> Option("urayagppd").collect { case x if x.length > end => x.substring(start, end) }
res15: Option[String] = None
