將XML數據文件寫入文本文件

-1

我在一個文件夾中有幾個XML文件。我想將XML數據文件寫入文本文件。但我不明白該怎麼做。 :(將XML數據文件寫入文本文件

示例XML文件：

<?xml version = "1.0"?> 
<note> 
<to> Mat </to> 
<from> Tim </from> 
<head> Black </head> 
<body> Yellow </body> 
</note>

這裏是我的代碼：

public class ReadXML extends DefaultHandler { 

Boolean noteTag = false; 
Boolean toTag = false; 
Boolean fromTag = false; 
Boolean headTag = false; 
Boolean bodyTag = false; 

static final String NOTE = "note"; 
static final String TO = "to"; 
static final String FROM = "from"; 
static final String HEAD = "head"; 
static final String BODY = "body"; 

public void read() throws Exception { 

    SAXParser saxParser = SAXParserFactory.newInstance().newSAXParser(); 

    File folder = new File("D:\\Source Code\\NetBeans\\Java\\BGPU\\ParseXMLFile"); 
    File[] listOfFiles = folder.listFiles(); 

    for (File file : listOfFiles) { 
     if (file.isFile() && file.getName().endsWith(".xml")) { 
      saxParser.parse(file, this); 
      System.out.println(); 
     } 
    } 
} 

@Override 
public void startElement(String uri, String localName, String qName, Attributes attributes) throws SAXException { 

    if (qName.equalsIgnoreCase(NOTE)) { 
     noteTag = true; 
    } 

    if (qName.equalsIgnoreCase(TO)) { 
     toTag = true; 
    } 

    if (qName.equalsIgnoreCase(FROM)) { 
     fromTag = true; 
    } 

    if (qName.equalsIgnoreCase(HEAD)) { 
     headTag = true; 
    } 

    if (qName.equalsIgnoreCase(BODY)) { 
     bodyTag = true; 
    } 
} 

@Override 
public void characters(char ch[], int start, int length) throws SAXException { 

    if (fromTag.equals(true)) { 
     System.out.println("FROM: " + new String(ch, start, length)); 
     } 

    if (toTag.equals(true)) { 
     System.out.println("TO: " + new String(ch, start, length)); 
     toTag = false; 
    } 

    if (headTag.equals(true)) { 
     System.out.println("HEAD: " + new String(ch, start, length)); 
     headTag = false; 
    } 
} 

@Override 
public void endElement(String uri, String localName, String qName) throws SAXException { 

    if (qName.equalsIgnoreCase(NOTE)) { 
     noteTag = false; 
    } 

    if (qName.equalsIgnoreCase(TO)) { 
     toTag = false; 
    } 

    if (qName.equalsIgnoreCase(FROM)) { 
     fromTag = false; 
    } 

    if (qName.equalsIgnoreCase(HEAD)) { 
     headTag = false; 
    } 

    if (qName.equalsIgnoreCase(BODY)) { 
     bodyTag = false; 
    } 
} 

public void save(String filename) throws Exception { 

}

請幫我完成save（）方法

來源

2014-09-27 s0lus

特別任何理由使用SAX解析器的更清潔的方式？您可以將文件複製到任何文件而無需解析內容：http://stackoverflow.com/questions/106770/standard-concise-way-to-copy-a-file-in-java – 2014-09-27 05:45:52

我只使用SAX，如果文件太大而不適合內存。 – Adam 2014-09-27 06:01:29

我建議不使用SAX，這裏是一個使用XPath的例子，你可以使用java.util.FileWriter寫入一個文件。

public class Test { 
    public static void main(String[] args) throws Exception { 
     BufferedWriter writer = new BufferedWriter(new FileWriter("output.txt")); 

     DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance(); 
     DocumentBuilder builder = factory.newDocumentBuilder(); 
     Document document = builder.parse("test.xml"); 
     XPathFactory xPathFactory = XPathFactory.newInstance(); 
     writer.append(xPathFactory.newXPath().compile("//note/to").evaluate(document)); 
     writer.newLine(); 
     writer.append(xPathFactory.newXPath().compile("//note/from").evaluate(document)); 
     writer.newLine(); 
     writer.append(xPathFactory.newXPath().compile("//note/head").evaluate(document)); 
     writer.newLine(); 
     writer.append(xPathFactory.newXPath().compile("//note/body").evaluate(document)); 
     writer.newLine(); 
     writer.close(); 
    } 
}

如果你真的必須使用SAX，這裏是做

public class ReadXml { 

    public static void main(String[] args) throws Exception { 
     new ReadXml().read(); 
    } 

    public void read() throws Exception { 

     SAXParser saxParser = SAXParserFactory.newInstance().newSAXParser(); 

     File folder = new File("."); 
     File[] listOfFiles = folder.listFiles(); 

     for (File file : listOfFiles) { 
      if (file.isFile() && file.getName().endsWith(".xml")) { 

       Handler handler = new Handler(); 
       saxParser.parse(file, handler); 
       save(handler, file.getName() + ".txt"); 
      } 
     } 
    } 

    private void save(Handler handler, String filename) throws IOException { 
     BufferedWriter writer = new BufferedWriter(new FileWriter(filename)); 
     writer.append(handler.getFrom()); 
     writer.newLine(); 
     writer.append(handler.getTo()); 
     writer.newLine(); 
     writer.append(handler.getHead()); 
     writer.newLine(); 
     writer.append(handler.getBody()); 
     writer.newLine(); 
     writer.close(); 
    } 

    private class Handler extends DefaultHandler { 

     private StringBuilder content; 
     private String to; 
     private String from; 
     private String body; 
     private String head; 

     @Override 
     public void startElement(String uri, String localName, String qName, 
       Attributes attributes) throws SAXException { 

      content = new StringBuilder(); 
     } 

     @Override 
     public void characters(char ch[], int start, int length) 
       throws SAXException { 
      content.append(ch, start, length); 
     } 

     @Override 
     public void endElement(String uri, String localName, String qName) 
       throws SAXException { 

      if ("to".equals(qName)) { 
       to = content.toString(); 
      } else if ("from".equals(qName)) { 
       from = content.toString(); 
      } else if ("body".equals(qName)) { 
       body = content.toString(); 
      } else if ("head".equals(qName)) { 
       head = content.toString(); 
      } 
     } 

     public String getTo() { 
      return to; 
     } 

     public String getFrom() { 
      return from; 
     } 

     public String getBody() { 
      return body; 
     } 

     public String getHead() { 
      return head; 
     } 

    } 
}

來源

2014-09-27 06:11:21 Adam

將XML數據文件寫入文本文件

回答

相關問題