Question

我有一个像这样的xml：

<Message xmlns="uri_of_message">
  <VendorId>1234</VendorId>
  <SequenceNumber>1</SequenceNumber>
  ...other important headers...
  <Data>
    <Functions xmlns="uri_of_functions_subxml">
      <Function1 attr="sth">
        <Info>Some_Info</Info>
      </Function1>
      <Function2>
        <Info>Some_Info</Info>
      </Function2>
      ...Functions n...
    </Functions>
  </Data>
</Message>

我需要提取内部xml

<Functions xmlns="uri_of_functions_subxml">
  <Function1 attr="sth">
    <Info>Some_Info</Info>
  </Function1>
  <Function2>
    <Info>Some_Info</Info>
  </Function2>
  ...Functions n...
</Functions>

我首先尝试使用字符方法获取内部xml：

 public void startElement(String uri, String localName, String tagName, Attributes attributes) throws SAXException {
if (tagName.equalsIgnoreCase("Data")){
  buffer = new StringBuffer();}
}
public void characters(char[] ch, int start, int length) throws SAXException {
  if (buffer != null) {
     buffer.append(new String(ch, start, length).trim());
  }
}
public void endElement(String uri, String localName, String tagName) throws SAXException {
if (tagName.equalsIgnoreCase("Data")){
innerXML = buffer.toString().trim();
}

但后来我意识到字符方法没有正确收集xml，它可能会拒绝像“＆lt;”，“＆gt;”这样的特殊字符。

下面的链接包含相同的问题，但答案不适用于我，因为外部xml必须作为握手信号处理，内部xml必须以完全不同的方式处理。

Java XML parsing: taking inner XML using SAX

我唯一需要的是正确收集内部xml。但是，怎么做呢？提前谢谢..

Answer 1

SAX似乎不是这项工作的最佳选择，无论如何尝试

    SAXParser p = SAXParserFactory.newInstance().newSAXParser();
    XMLReader filter = new XMLFilterImpl(p.getXMLReader()) {
        private boolean inFunctions;

        @Override
        public void startElement(String uri, String localName, String qName, Attributes atts) throws SAXException {
            if (!inFunctions && qName.equals("Functions")) {
                inFunctions = true;
            }
            if (inFunctions) {
                super.startElement(uri, localName, qName, atts);
            } else {
                qName.equals("Functions");
            }
        }

        @Override
        public void endElement(String uri, String localName, String qName) throws SAXException {
            if (inFunctions) {
                super.endElement(uri, localName, qName);
                if (qName.equals("Functions")) {
                    inFunctions = false;
                }
            }
        }

        @Override
        public void characters(char[] ch, int start, int length) throws SAXException {
            if (inFunctions) {
                super.characters(ch, start, length);
            }
        }
    };
    Transformer t = TransformerFactory.newInstance().newTransformer();
    Source source = new SAXSource(filter, new InputSource(new FileInputStream("1.xml")));
    Result result = new StreamResult(System.out);
    t.transform(source, result);
}

输出

<?xml version="1.0" encoding="UTF-8"?><Functions xmlns="uri_of_functions_subxml">
      <Function1 attr="sth">
        <Info>Some_Info</Info>
      </Function1>
      <Function2>
        <Info>Some_Info</Info>
      </Function2>
    </Functions>

Official Tutorial

使用sax获取内部xml

1 个答案: