Question

我有一个我想要转换为xml的字符串列表。我如何使用DOM？我不确定如何使用字符串创建Document以用作Source来生成XML。

有人能告诉我生成文档的示例代码吗？例如，XML就像：

ArrayList<String> fruits

<Fruits>
  <fruit>Apple<fruit>
  <fruit>Grape<fruit>
<Fruits>

我认为代码是：

TransformerFactory transFact = TransformerFactory.newInstance();
Transformer serializer = transFact.newTransformer();
Properties props = new Properties();
props.put("method", "xml");
props.put("indent", "yes");
serializer.setOutputProperties(props);
Source source = new DOMSource(document); //need to create a document
Result result = new StreamResult(PrintWriter);

Answer 1

    ArrayList<String> a = new ArrayList<String>();
    a.add("apple"); a.add("mango");

    DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance();
    DocumentBuilder docBuilder = docFactory.newDocumentBuilder();

    Document doc = docBuilder.newDocument();
    Element root = doc.createElement("Fruits");
    doc.appendChild(root);

    for(String name: a){
        Element fruit = doc.createElement("fruit");
        fruit.appendChild(doc.createTextNode(name)); 
        root.appendChild(fruit);
    }
    TransformerFactory transformerFactory = TransformerFactory.newInstance();
    Transformer transformer = transformerFactory.newTransformer();
    DOMSource source = new DOMSource(doc);
    StreamResult result = new StreamResult(new File("C:\\file.xml"));

    transformer.transform(source, result);

Answer 2

我会使用JAXB：

@XmlRootElement(name = "Fruits")
@XmlAccessorType(XmlAccessType.FIELD)
public static class Fruits
{
    @XmlElement(name = "fruit")
    public List<String> fruit = new ArrayList<String>();
}

public static void main(String[] args) throws Exception
{
    Fruits fruits = new Fruits();
    fruits.fruit.add("Apple");
    fruits.fruit.add("Grape");

    TransformerFactory transFact = TransformerFactory.newInstance();
    Transformer serializer = transFact.newTransformer();
    Properties props = new Properties();
    props.put("method", "xml");
    props.put("indent", "yes");
    serializer.setOutputProperties(props);
    Source source = new JAXBSource(JAXBContext.newInstance(Fruits.class), fruits);
    Result result = new StreamResult(System.out);
    serializer.transform(source, result);
}

Answer 3

如果你真的想使用“fruits”列表中的字符串创建一个Document作为Source：

    StringBuilder sb  = new StringBuilder();
    for(String s : fruits) {
        sb.append(s).append('\n');
    }
    DocumentBuilderFactory f = DocumentBuilderFactory.newInstance();
    Document doc = f.newDocumentBuilder().parse(new InputSource(new StringReader(sb.toString())));

从字符串创建XML

3 个答案: