1
我尝试迭代XML文件以获取特定类型的所有元素<Card>
。这种类型也可以具有可选的属性。如何迭代可以具有可选属性的XML元素?
XML示例:
<Lesson>
<Description>Description</Description>
<Card learnedTimestamp="1234567", isRepeatedByTyping="true", batch="5">
<FrontSide>Foo1</FrontSide>
<ReverseSide>Bar1</ReverseSide>
<InformationLine>Info1</InformationLine>
</Card>
<Card>
<FrontSide>Foo2</FrontSide>
<ReverseSide>Bar2</ReverseSide>
<InformationLine>Info2</InformationLine>
</Card>
</Lesson>
Scala代码:
class XMLParser(fqFileName: String) {
val pauDoc: Elem = XML.loadFile(fqFileName)
def printXMLFile() = {
var cardCount = 0
val lesson = (pauDoc \\ "Lesson")
for(val card <- lesson \\ "Card"){
cardCount = cardCount + 1
println("Card No " + cardCount)
val frontSide = (card \\ "FrontSide").text
println("FrontSide Value: " + frontSide)
val reverseSide = (card \\ "ReverseSide").text
println("ReverseSide Value.text: " + reverseSide)
val infoLine = (card \\ "InformationLine").text
println("InformationLine Value: " + infoLine)
}
}
}
但这个代码,如果<Card>
元素没有任何属性才有效。有人知道我如何处理属性,特别是如果它们是可选的?
P.S .:这是我得到的例外。
Exception in thread "main" org.xml.sax.SAXParseException: Element type "Card" must be followed by either attribute specifications, ">" or "/>".
at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.createSAXParseException(Unknown Source)
at com.sun.org.apache.xerces.internal.util.ErrorHandlerWrapper.fatalError(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLErrorReporter.reportError(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLScanner.reportFatalError(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.seekCloseOfStartTag(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanStartElement(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl$FragmentContentDriver.next(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLDocumentScannerImpl.next(Unknown Source)
at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanDocument(Unknown Source)
at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(Unknown Source)
at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(Unknown Source)
at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(Unknown Source)
at com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.parse(Unknown Source)
at com.sun.org.apache.xerces.internal.jaxp.SAXParserImpl$JAXPSAXParser.parse(Unknown Source)
at javax.xml.parsers.SAXParser.parse(Unknown Source)
at scala.xml.factory.XMLLoader$class.loadXML(XMLLoader.scala:40)
at scala.xml.XML$.loadXML(XML.scala:40)
at scala.xml.factory.XMLLoader$class.loadFile(XMLLoader.scala:49)
at scala.xml.XML$.loadFile(XML.scala:40)
at de.htwg_konstanz.ecardman.common.XMLParser.<init>(XMLParser.scala:7)
at de.htwg_konstanz.ecardman.common.XMLParserMain$.main(XMLParserMain.scala:23)
at de.htwg_konstanz.ecardman.common.XMLParserMain.main(XMLParserMain.scala)
是的,这就是我所期望的,如果我使用loadFile。但是我也需要访问元素的属性(如果可用)。你知道如何做到这一点? –
DaHanz
2012-01-13 09:30:21
omg,抱歉一切正常,问题是属性之间的“,”。 :/ – DaHanz 2012-01-13 10:14:50