Creating XML from Strings

Question

I have a list of strings that i want to convert to xml. How would I do so using DOM? I'm not sure how to create a Document using the strings to use as a Source to generate the XML.

Can anyone show me a sample code to generate a Document? For example the XML would be like:

ArrayList<String> fruits

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

I think the code would be:

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);

Robin Chander · Accepted Answer · 2012-11-19 15:55:47Z

3

    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);

edited Nov 19, 2012 at 15:55

answered Nov 19, 2012 at 15:08

Robin Chander

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2 Comments

Dan Over a year ago

the result is not what I expected. The output is <Fruits><fruit>applegrapepeanut</fruit></Fruits>. I want each fruit to be in a separate <fruit> tag

Robin Chander Over a year ago

you might be doing something wrong. Look at my updated answer, I have removed the comments and added the iterator code.

artbristol · Accepted Answer · 2012-11-19 15:41:57Z

1

I would use JAXB for this:

@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);
}

edited Nov 19, 2012 at 15:41

answered Nov 19, 2012 at 15:07

artbristol

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2 Comments

bdoughan Over a year ago

Using a JAXB (JSR-222) implementation you could just use a Marshaller to write the object to the desired XML output instead of using a Transformer. The approach you give is useful when applying a stylesheet to an JAXB source: blog.bdoughan.com/2012/11/…

artbristol Over a year ago

@BlaiseDoughan indeed, I was only copying the OP's snippet. I would have just used JAXB.marshal(fruits, System.out) otherwise.

Evgeniy Dorofeev · Accepted Answer · 2012-11-19 15:10:31Z

0

if you really want to create a Document using strings in "fruits" list as a 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())));

answered Nov 19, 2012 at 15:10

Evgeniy Dorofeev

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