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我有一个斯卡拉理解unapplySeq
object radExtractor{
def unapplySeq(row:HtmlTableRow):Option[List[String]]={
val lista = (for{
a<-row.getByXPath("td/span/a")
ah= a.asInstanceOf[DomNode]
if(ah.getFirstChild!=null)
} yield a.asInstanceOf[DomNode].getFirstChild.toString).toList
lista match{
case Nil=>None
case l @ List(duns,companyname,address,city,postal,_bs,orgnummer, _*) =>Some(l)
case _ =>println("WTF");None
}
}
}
,我想在一个列表理解一样使用它:
val toReturn = for{
rad<-rader
val radExtractor(duns,companyname,address,city,postal,_,orgnummer,_*)=rad
} yield Something(duns,companyname,address,city,postal,orgnummer)
但是,当“拉德”的“雷德”失败,因为提取返回None
我收到MatchError
。
是不是理解提取器应该处理/忽略None
案件或我只是错过了什么?
我能做
val toReturn = rader.collect{case radExtractor(duns,companyname,address,city,postal,_,orgnummer, _*)=>
Something(companyname=companyname,address=address,city=city,postalcode=postal,orgnummer=orgnummer,duns=duns.toInt)
}
但是,这不会被视为性感;) 谢谢