使用Scala将XML转换为系统属性

使用Scala，转换的最佳方法是什么

<myapp>
  <username>bill</username>
  <password>secret123</password>
  <background>#FFFFFF</background>
</myapp>

---

假设转换器附加到sys.props？

支持XML格式，虽然不是您描述的格式。

谢谢，我如何处理更深层次的条目，例如william，以解析为myapp.username.secondary=william？您的XML是否会被修复（即，可能只存在少量定义良好的标记）？如果是这样的话，那么只做固定的XPath表达式可能会更容易。例如，val secondary=e\\“secondary”；如果（！secondary.isEmpty）sp+=（“myapp.username.secondary”，secondary.text）理想情况下希望它是一种开放格式，但这也很好：）您可以将我在答案中所做的调整为开放格式-您需要使其递归。但基本概念/API是相同的。

scala> val s = "<myapp><username>bill</username><password>secret123</password><background>#FFFFFF</background></myapp>"
s: java.lang.String = <myapp><username>bill</username><password>secret123</password><background>#FFFFFF</background></myapp>

scala> val e = xml.XML.loadString(s)
e: scala.xml.Elem = <myapp><username>bill</username><password>secret123</password><background>#FFFFFF</background></myapp>

scala> val sp = new sys.SystemProperties
sp: scala.sys.SystemProperties = 
Map(env.emacs -> "", java.runtime.name -> Java(TM) SE Runtime Environment, ....)

scala> sp ++= e.child.map(n => (e.label + "." + n.label, n.text))
res11: sp.type = 
Map(env.emacs -> "", java.runtime.name -> Java(TM) SE Runtime Environment, ...)

scala> val s = "<myapp><username>bill</username><password>secret123</password><background>#FFFFFF</background></myapp>"
s: java.lang.String = <myapp><username>bill</username><password>secret123</password><background>#FFFFFF</background></myapp>

scala> val e = xml.XML.loadString(s)
e: scala.xml.Elem = <myapp><username>bill</username><password>secret123</password><background>#FFFFFF</background></myapp>

scala> val sp = new sys.SystemProperties
sp: scala.sys.SystemProperties = 
Map(env.emacs -> "", java.runtime.name -> Java(TM) SE Runtime Environment, ....)

scala> sp ++= e.child.map(n => (e.label + "." + n.label, n.text))
res11: sp.type = 
Map(env.emacs -> "", java.runtime.name -> Java(TM) SE Runtime Environment, ...)

scala> val p = java.lang.System.getProperties
p: java.util.Properties = 
{env.emacs=, java.runtime.name=Java(TM) SE Runtime Environment,...}

scala> import collection.JavaConversions._
import collection.JavaConversions._

scala> p filter { case (k, v) => k.startsWith("myapp") } \
          foreach { case (k,v) => println(k + "=" + v) }
myapp.password=secret123
myapp.background=#FFFFFF
myapp.username=bill