如何根据选项添加或不添加 XML 属性? [英] How do I add an XML attribute, or not, depending on an Option?

问题描述

我编写了一个 makeMsg 函数，但我不喜欢它 - 基于 Option.isDefined 进行区分似乎真的不符合 Scala 风格.你能做得更好吗?

I have written a makeMsg function but I don't like it - it just seems really un-Scala-ish to discriminate based on Option.isDefined. Can you make it better?

scala> def makeMsg(t: Option[String]) = 
     | if (t.isDefined) <msg text={t.get} /> else <msg />
makeMsg: (t: Option[String])scala.xml.Elem

scala> makeMsg(Some("hello"))
res0: scala.xml.Elem = <msg text="hello"></msg>

scala> makeMsg(None)
res1: scala.xml.Elem = <msg></msg>

推荐答案

你可以试试这个:

def makeMsg(t: Option[String]) = <msg text={t orNull} />

如果属性值为 null - 它不会被添加到元素中.

if attribute value is null - it will not be added to the element.

更好！如果您将添加此隐式转换:

Even better! If you will add this implicit convertion:

import xml.Text
implicit def optStrToOptText(opt: Option[String]) = opt map Text

你可以像这样使用t:

def makeMsg(t: Option[String]) = <msg text={t} />

这是 REPL 会话:

Here is REPL session:

scala> import xml.Text
import xml.Text

scala> implicit def optStrToOptText(opt: Option[String]) = opt map Text
optStrToOptText: (opt: Option[String])Option[scala.xml.Text]

scala> def makeMsg(t: Option[String]) = <msg text={t} />
makeMsg: (t: Option[String])scala.xml.Elem

scala> makeMsg(Some("hello"))
res1: scala.xml.Elem = <msg text="hello"></msg>

scala> makeMsg(None)
res2: scala.xml.Elem = <msg ></msg>

这是可行的，因为 scala.xml.UnprefixedAttribute 具有接受 Option[Seq[Node]] 作为值的构造函数.

This works because scala.xml.UnprefixedAttribute has constructor that accepts Option[Seq[Node]] as value.

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