Question

我有一个文本文件：

engMusic
  Anastasia-Te_Iert.mp3
  Calvin_Harris_and_Alesso_ft_Hurts-Under_Control.mp3
  Oceana-Endless_Summer_Remix.mp3
  The_Wanted-Show_Me_Love.mp3
rusMusic
  basta-feat-smoki-mo-kamennye-cvety-solovey.su.mp3
  maks-barskih-zdes-i-seychas.mp3

我使用SAX解析此文件，我想要像这样获取XML文件

<Music>
<CATALOG_NAME>engMusic</CATALOG_NAME>
    <FILE_NAME>Anastasia-Te_Iert.mp3</FILE_NAME>
    <FILE_NAME>Calvin_Harris_and_Alesso_ft_Hurts-Under_Control.mp3</FILE_NAME>
    <FILE_NAME>Oceana-Endless_Summer_Remix.mp3</FILE_NAME>
    <FILE_NAME>The_Wanted-Show_Me_Love.mp3</FILE_NAME>
<CATALOG_NAME>rusMusic</CATALOG_NAME>
    <FILE_NAME>basta-feat-smoki-mo-kamennye-cvety-solovey.su.mp3</FILE_NAME>
    <FILE_NAME>maks-barskih-zdes-i-seychas.mp3</FILE_NAME>
</Music>

但我有一个问题。我不知道如何选择子文件夹并为“rusMusic”分配标签。实际上我得到了这个输出：

<Music>
<CATALOG_NAME>engMusic</CATALOG_NAME>
    <FILE_NAME>Anastasia-Te_Iert.mp3</FILE_NAME>
    <FILE_NAME>Calvin_Harris_and_Alesso_ft_Hurts-Under_Control.mp3</FILE_NAME>
    <FILE_NAME>Oceana-Endless_Summer_Remix.mp3</FILE_NAME>
    <FILE_NAME>The_Wanted-Show_Me_Love.mp3</FILE_NAME>
**<FILE_NAME>rusMusic</FILE_NAME>** // Actually should be <CATALOG_NAME>rusMusic</CATALOG_NAME>
    <FILE_NAME>basta-feat-smoki-mo-kamennye-cvety-solovey.su.mp3</FILE_NAME>
    <FILE_NAME>maks-barskih-zdes-i-seychas.mp3</FILE_NAME>
</Music>

我的代码：

public class ConvertToXML {

BufferedReader in;
StreamResult out;

TransformerHandler th;
AttributesImpl atts;

public void convertToXml() {

    try {
        in = new BufferedReader(new FileReader("content.txt"));
        out = new StreamResult("dir.xml");
        initXML();

        String str;
        ArrayList<String> content = new ArrayList<>();

        while ((str = in.readLine()) != null) {
            content.add(str);
        }
        process(content);

        in.close();
        writeXML();
    } catch (Exception ex) {
        ex.printStackTrace();
    }
} 

private void initXML() throws ParserConfigurationException, TransformerConfigurationException, SAXException {

    SAXTransformerFactory tf = (SAXTransformerFactory) SAXTransformerFactory.newInstance();
    th = tf.newTransformerHandler();
    Transformer transformer = th.getTransformer();
    transformer.setOutputProperty(OutputKeys.ENCODING, "UTF-8");
    transformer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "4");
    transformer.setOutputProperty(OutputKeys.INDENT, "yes");
    th.setResult(out);
    th.startDocument();
    atts = new AttributesImpl();
    th.startElement("", "", "Music", atts);

}

private void process(ArrayList<String> elements) throws SAXException {

    atts.clear();

    th.startElement("", "", "CATALOG_NAME", atts);
    th.characters(elements.get(0).toCharArray(), 0, elements.get(0).length());
    th.endElement("", "", "CATALOG_NAME");

    for (int i = 1; i < elements.size(); i++) {
        th.startElement("", "", "FILE_NAME", atts);
        th.characters(elements.get(i).toCharArray(), 0, elements.get(i).length());
        th.endElement("", "", "FILE_NAME"); 
    } 
}

private void writeXML() throws TransformerConfigurationException, TransformerException, SAXException {  
    th.endElement("", "", "Music");
    th.endDocument();
}

}

Answer 1

我选择这样的事情（没有尝试过，直接在这里输入）

在convertToXml()

while ((str = in.readLine()) != null) {
    process(str);
}

process()现在应该接受String

private void process(String row) throws SAXException {

    if (row.startsWith("  ")) {
        th.startElement("", "", "FILE_NAME", atts);
        th.characters(row.toCharArray(), 0, row.length());
        th.endElement("", "", "FILE_NAME"); 
    } else {
        th.startElement("", "", "CATALOG_NAME", atts);
        th.characters(row.toCharArray(), 0, row.length());
        th.startElement("", "", "CATALOG_NAME", atts);
    }

}

这可能不会为自己工作，但你明白了。只测试行是否从文件中的任何内容开始（2或4个空格，TAB字符，...），在这种情况下它是FILE_NAME，否则它是CATALOG_NAME。

xml文件的结构

1 个答案: