scala – 是否有可能从不同的特征中链接方法?
发布时间:2020-12-16 18:59:08 所属栏目:安全 来源:网络整理
导读:我有以下代码: class Parameterizable{ var map: Map[String,String] = new scala.collection.immutable.HashMap() def put(entry: Tuple2[String,String]) = { map = map + entry; this }}class Query() extends Parameterizable{ override def toString =
|
我有以下代码:
class Parameterizable{
var map: Map[String,String] = new scala.collection.immutable.HashMap()
def put(entry: Tuple2[String,String]) = {
map = map + entry; this
}
}
class Query() extends Parameterizable{
override def toString = {
map.isEmpty match{
case true => ""
case false => "?" + map.map{case (key,value) => key + "=" + value}.mkString("&")
}
}
}
trait PageParameter extends Parameterizable{
def page(page: Int) = put(("page" -> page.toString))
def pageSize(pageSize: Int) = put(("pagesize" -> pageSize.toString))
}
trait DateParameter extends Parameterizable{
def fromDate(date: java.util.Date) = put(("fromdate" -> (date.getTime()/1000L).toString()))
def toDate(date: java.util.Date) = put(("todate" -> (date.getTime()/1000L).toString()))
}
//and other similar traits
我想做的事情如下: class ExtendedQuery extends Query with PageParameter with DateParameter val query = new ExtendedQuery query.page(4).pageSize(5).fromDate(new java.util.Date) 要么: query.and().page(4).and().pageSize(5).and().fromDate(new java.util.Date) Scala有可能吗? 解决方法
您可以将方法声明为返回this.type,然后从中返回:
trait PageParameter extends Parameterizable{
def page(page: Int) : this.type = { put(("page" -> page.toString)); this }
def pageSize(pageSize: Int): this.type = { put(("pagesize" -> pageSize.toString)); this }
}
在使用现场,您可以根据需要链接呼叫.看这个例子: scala> trait Wibble {
| def foo : this.type = { println("foo"); this }
| }
defined trait Wibble
scala> trait Wobble extends Wibble {
| def bar: this.type = { println("bar"); this }
| }
defined trait Wobble
scala> trait Wubble extends Wibble {
| def baz: this.type = { println("baz"); this }
| }
defined trait Wubble
现在我可以测试一下 scala> new Wibble with Wobble with Wubble res0: java.lang.Object with Wibble with Wobble with Wubble = $anon$1@937e20 scala> res0.bar.baz.foo bar baz foo res1: res0.type = $anon$1@937e20 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |
