Java XML 获取属性 [英] Java XML getAttribute

问题描述

我正在尝试从我的 XML 文档中获取一个属性 id (fileID) 以用作我的 XML 拆分的文件名.拆分工作我只需要提取 fileID 用作名称.

I'm trying to get an attribute id (fileID) from my XML document to use as the filename for my XML split. The split works I just need to extract the fileID to use as the name.

我可以用它来帮助解决这个问题.

I could use this as help on this.

这是我的xml文档

<root>
 <envelope fileID="000152OP.XML">
   <record id="850">
   </record>
</envelope>
<envelope fileID="000153OP.XML">
  <record id="850">
  </record>
</envelope>
<envelope fileID="000154OP.XML">
  <record id="850">
  </record>
</envelope>
</root>

这是我的 Java 代码 [已编辑] 我现在可以读取该属性，但它不会创建最后一个 xml 文件.因此，在我的示例中，它使用正确的名称创建前 2 个文件，但未创建最后一个文件 ID000154OP.XML".

And here's my Java code I can read the attribute now but it doesn't create the last xml file. So in my example it create the first 2 files with the correct name but last fileID "000154OP.XML" isn't created.

    public static void splitXMLFile (String file) throws Exception {         
    String[] temp;
    String[] temp2;
    String[] temp3;
    String[] temp4;
    String[] temp5;
    String[] temp6;
    File input = new File(file);         
    DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();         
    Document doc = dbf.newDocumentBuilder().parse(input);
    XPath xpath = XPathFactory.newInstance().newXPath();          
    NodeList nodes = (NodeList) xpath.evaluate("//root/envelope", doc, XPathConstants.NODESET);          
    int itemsPerFile = 1;         

    Node staff = doc.getElementsByTagName("envelope").item(0);

    NamedNodeMap attr = staff.getAttributes();
    Node nodeAttr = attr.getNamedItem("fileID");
    String node = nodeAttr.toString();
    temp = node.split("=");
    temp2 = temp[1].split("^\"");
    temp3 = temp2[1].split("\\.");

    Document currentDoc = dbf.newDocumentBuilder().newDocument();         
    Node rootNode = currentDoc.createElement("root");   
    File currentFile = new File("C:\\XMLFiles\\" + temp3[0]+ ".xml"); 

    for (int i=1; i <= nodes.getLength(); i++) {             
        Node imported = currentDoc.importNode(nodes.item(i-1), true);             
        rootNode.appendChild(imported); 

        Node staff2 = doc.getElementsByTagName("envelope").item(i);
        NamedNodeMap attr2 = staff2.getAttributes();
        Node nodeAttr2 = attr2.getNamedItem("fileID");
        String node2 = nodeAttr2.toString();
        temp4 = node2.split("=");
        temp5 = temp4[1].split("^\"");
        temp6 = temp5[1].split("\\.");

        if (i % itemsPerFile == 0) { 

            writeToFile(rootNode, currentFile);                  
            rootNode = currentDoc.createElement("root");    
            currentFile = new File("C:\\XMLFiles\\" + temp6[0]+".xml");


        }         
    }          
    writeToFile(rootNode, currentFile);     
}    

 private static void writeToFile(Node node, File file) throws Exception {         
     Transformer transformer = TransformerFactory.newInstance().newTransformer();         
     transformer.transform(new DOMSource(node), new StreamResult(new FileWriter(file)));     
 } 

推荐答案

或许你可以试试下面的xpath:

Perhaps you can try the following xpath:

//root/envelope/record/@id

如果您使用的 XPath 库不支持整个 XPath 集，您可以尝试优秀的库 jaxen

If the XPath library you're using does not support the entire XPath set you can try the excellent library jaxen

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