在我阅读了xml.Utility.unescape的SDK文档后,我认为它是xml.Utility.escape的反向操作,但它似乎根本没有做任何事情：
scala> xml.Utility.escape("& <")
res0: String = & <
scala> val sb = new StringBuilder
sb: StringBuilder =
scala> xml.Utility.unescape("& <",sb)
res1: StringBuilder = null
scala> sb.toString
res2: String = ""
如何正确使用xml.Utility.unescape？
解决方法
我查看了unescape文档,我非常失望……
Appends unescaped string to s,amp becomes &,lt becomes < etc..
returns null if ref was not a predefined entity.
因此,看起来unescape意味着将单个字符附加到StringBuilder.它认为“& amp;& lt;”是“不是预定义的实体”所以它为你的res1返回null.
REPL中的一些测试：
scala> unescape("amp",sb)
res9: StringBuilder = &
scala> unescape("lt",sb)
res10: StringBuilder = &<
scala> unescape(" ",sb)
res11: StringBuilder = null
