Question

我想借助java代码将一个xml转换为另一个xml。我想给一个xml作为输入文件，并期望另一个xml文件作为输出。我们如何在java的帮助下做这件事？任何人都可以向我提出建议。

Input.xml文件

<Order OrderNo=”1234567890”>

 <OrderLines>

  <OrderLine PrimeLineNo=”1” SubLineNo=”1”/>

  <OrderLine PrimeLineNo=”2” SubLineNo=”1”/>

 </OrderLines>

</Order>

output.xml文件我需要像这样输出这个文件

<Order OrderName="1234567890">

 <OrderLines MaxOrderNumbers=”2”>

  <OrderLine LineNumber="1" SubLineNumber="1"/>

  <OrderLine LineNumber ="2" SubLineNumber ="1"/>

 </OrderLines>

</Order>

但我已经在java的帮助下尝试了下面的代码。

public class XmlToXml {

public static void main(String[] args) {

    final String xmlStr ="<Order OrderNo=\"1234567890\"><OrderLines><OrderLine PrimeLineNo=\"1\"                     SubLineNo=\"1\"/><OrderLine PrimeLineNo=\"2\" SubLineNo=\"1\"/></OrderLines></Order>";

    Document doc = convertStringToDocument(xmlStr);

    String str = convertDocumentToString(doc);

      System.out.println(str);
}

private static String convertDocumentToString(Document doc) {

    TransformerFactory tf = TransformerFactory.newInstance();

    Transformer transformer;
    try {
        transformer = tf.newTransformer();

         System.out.println(transformer.getParameter("xmlStr"));

        StringWriter writer = new StringWriter();

        writer.append("MaxOrderNumbers");

        transformer.transform(new DOMSource(doc), new StreamResult(writer));

        String output = writer.getBuffer().toString();

        return output;

    } catch (TransformerException e) {

        e.printStackTrace();
    }

    return null;
}

private static Document convertStringToDocument(String xmlStr) {

    DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();  
    DocumentBuilder builder;  
    try 
    {  
        builder = factory.newDocumentBuilder(); 

        Document doc = builder.parse(new InputSource( new StringReader( xmlStr ) ) );

        return doc;

    } catch (Exception e) { 

        e.printStackTrace();  
    } 
    return null;
}
}

Answer 1

解决这个问题的方法是在XSLT中定义转换规则。

默认情况下，您需要一个可以复制内容的规则：

<xsl:template match="*|@*">
 <xsl:copy>
  <xsl:apply-templates select="*|@*"/>
 </xsl:copy>
</xsl:template>

然后您需要进一步的规则来定义您想要进行的更改：

<xsl:template match="@OrderNo">
 <xsl:attribute name="OrderName"><xsl:value-of select="."/></xsl:attribute>
</xsl:template>

和其他重命名规则类似，加上

<xsl:template match="OrderLines">
 <OrderLines MaxOrderLines="{count(*)}">
  <xsl:apply-templates select="*|@*"/>
 </OrderLines>
</xsl:template>

然后，将这些规则组合到样式表中，您可以使用与您已经完成的代码非常相似的代码从Java代码运行转换：

TransformerFactory tf = TransformerFactory.newInstance();
Templates t = tf.newTemplates(new StreamSource(new File("stylesheet.xsl")));
StringWriter writer = new StringWriter();
t.newTransformer().transform(
 new StreamSource(new File("input.xml")), 
 new StreamResult(writer));

Answer 2

如果您所做的只是更改标记和属性的文本，而不是XML文件本身的结构，那么您可以简单地将其作为文本文件读取，并使用Regex将一个字符串的所有实例与另一个字符串交换。以下代码可以帮助您入门：

//read in the file one line at a time
BufferedReader in = new BufferedReader(new FileReader("input.xml"));
FileOutputStream fos = new FileOutputStream("output.xml");
try {

    StringBuilder sb = new StringBuilder();
    String l = "";
    String nl = System.getProperty("line.separator");

    //get next line, assign it to l, and make sure it exists. The line after the last in the file will always be null
    while((l = in.readLine()) != null) {
        //replace the desired substrings of each line
        if(l.contains("PrimeLineNo")) {
            l = l.replaceAll("PrimeLineNo", "LineNumber");
        }
        if(l.contains("SubLineNo")) {
            l = l.replaceAll("SubLineNo", "SubLineNumber");
        }
        //add the line to a string buffer to be written to file later
        sb.append(l + nl);
    }   

    //after we have gone through the entire input file, write it to the output file
    fos.write(sb.toString().getBytes());
}
catch(Exception e) {
    System.out.println("An unknown error occurred");
}
finally {
    //ensure the streams will always close
    in.close();
    fos.close();
}

使用Java转换XML文件

2 个答案: