如何在没有文字元素名称的情况下在 Scala 中创建 XML 根节点? [英] How do I create an XML root node in Scala without a literal element name?

问题描述

我想创建一个这样的文档:

I'm looking to create a document like this:

<root/>

我可以以编程方式添加子项.理论上，它看起来像这样:

That I can add children to programatically. Theoretically, it would look like this:

val root_node_name = "root"
val doc = <{root_node_name}/>

但这似乎不起作用:

error: not found: value <

所以，我尝试的是这样的:

So, what I tried instead was this:

val root_node_name = "root"
val doc = new scala.xml.Elem(null, root_node_name, null, scala.xml.TopScope, null)  

那可以编译，但在运行时我得到这个空指针异常:

That compiles but at runtime I get this null pointer exception:

java.lang.NullPointerException
at scala.xml.Utility$.toXML(Utility.scala:201)
at scala.xml.Utility$$anonfun$sequenceToXML$2.apply(Utility.scala:235)
at scala.xml.Utility$$anonfun$sequenceToXML$2.apply(Utility.scala:235)
at scala.Iterator$class.foreach(Iterator.scala:414)
at scala.runtime.BoxedArray$AnyIterator.foreach(BoxedArray.scala:45)
at scala.Iterable$class.foreach(Iterable...

我使用的是 Scala 2.8.任何有关如何实现这一目标的示例?谢谢.

I'm using Scala 2.8. Any examples of how to pull this off? Thanks.

推荐答案

你应该传递属性的空列表 (scala.xml.Null)，如果你不想要任何孩子，你甚至不应该包括最后的论点.您需要一个空的子项列表，而不是一个恰好为 null 的子项.所以:

You should pass the empty list for attributes (scala.xml.Null) and if you don't want any children, you shouldn't even include the final argument. You want an empty list of children, not a single child that happens to be null. So:

scala> val root_node_name = "root"
root_node_name: java.lang.String = root

scala> val doc = new scala.xml.Elem(null, root_node_name, scala.xml.Null , scala.xml.TopScope)
doc: scala.xml.Elem = <root></root>

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