您似乎必须将XML文本中的属性值显式指定为字符串:
scala> val foo = 3
foo: Int = 3
scala> <bar id={ foo } />
<console>:10: error: overloaded method constructor UnprefixedAttribute with alternatives:
(key: String,value: Option[Seq[scala.xml.Node]],next: scala.xml.MetaData)scala.xml.UnprefixedAttribute <and>
(key: String,value: String,next: scala.xml.MetaData)scala.xml.UnprefixedAttribute <and>
(key: String,value: Seq[scala.xml.Node],next1: scala.xml.MetaData)scala.xml.UnprefixedAttribute
cannot be applied to (java.lang.String, Any, scala.xml.MetaData)
<bar id={ foo } />
^
scala> <bar id={ foo.toString } />
res16: scala.xml.Elem = <bar id="3"></bar>
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构造函数不能简单地接受Any参数并在其上调用toString吗?由于以下工作,
scala> <bar>{ foo }</bar>
res21: scala.xml.Elem = <bar>3</bar>
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这个API有一点点不对称.这有什么特别的原因吗?
小智 3
您可以为这种情况添加隐式转换:
scala> val foo = 3
foo: Int = 3
scala> implicit def anyToText( a: AnyVal ) = xml.Text( a.toString )
anyToText: (a: AnyVal)scala.xml.Text
scala> <bar id={foo}/>
res2: scala.xml.Elem = <bar id="3"></bar>
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