New to SAX Parser so need this BASIC

Question

if we do this in SAX Parser:

public class SAXXMLHandler extends DefaultHandler {

    private List<Employee> employees;
    private String tempVal;
    private Employee tempEmp;

    public SAXXMLHandler() {
        employees = new ArrayList<Employee>();
    }

    public List<Employee> getEmployees() {
        return employees;
    }

    // Event Handlers
    public void startElement(String uri, String localName, String qName,
            Attributes attributes) throws SAXException {
        // reset
        tempVal = "";
        if (qName.equalsIgnoreCase("employee")) {
            // create a new instance of employee
            tempEmp = new Employee();
        }
    }

    public void characters(char[] ch, int start, int length)
            throws SAXException {
        tempVal = new String(ch, start, length);
    }

    public void endElement(String uri, String localName, String qName)
            throws SAXException {
        if (qName.equalsIgnoreCase("employee")) {
            // add it to the list
            employees.add(tempEmp);
        } else if (qName.equalsIgnoreCase("id")) {
            tempEmp.setId(Integer.parseInt(tempVal));
        } else if (qName.equalsIgnoreCase("name")) {
            tempEmp.setName(tempVal);
        } else if (qName.equalsIgnoreCase("department")) {
            tempEmp.setDepartment(tempVal);
        } else if (qName.equalsIgnoreCase("type")) {
            tempEmp.setType(tempVal);
        } else if (qName.equalsIgnoreCase("email")) {
            tempEmp.setEmail(tempVal);
        }
    }
}

for this :

<employee>
        <id>2163</id>
        <name>Kumar</name>
        <department>Development</department>
        <type>Permanent</type>
        <email>kumar@tot.com</email>
</employee>

What we will do in SAX Parser for this :

<MyResource>
<Item>First</Item>
<Item>Second</Item>
</MyResource>

I am a newbie to SAX Parser.
Previously i had problems with DOM and PullParser for the same XML.
Isn't there any parser built for parsing this simple XML.

Raphaël · Accepted Answer · 2013-07-04T08:14:23.357

To parse an entry named MyResource that contains multiple entry of Item you can do something like that :

At first, initialize your variables inside the startDocument method to allow the reuse of your Handler :

private List<MyResource> resources;
private MyResource currentResource;
private StringBuilder stringBuilder;

public void startDocument() throws SAXException {
    map = null;
    employees = new ArrayList<Employee>();
    stringBuilder = new StringBuilder();
}

Detect inside of startElement when a MyResource is starting. The tag name will usually be stored inside the qName argument. When a MyResource is starting, you may want to create an instance of MyResource as a temporary variable. You will feed it until its end tag is reached.

public void startElement(String uri, String localName, String qName, Attributes attributes) throws SAXException {
    if ("MyResource".equals(qName)) {
        currentResource = new MyResource();
    }
    stringBuilder.setLength(0); // Reset the string builder
}

The characters method is required to read the content of each tag. Use StringBuilder to read characters. SAX may call more than once the characters method for each tag :

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

Inside the endElement, create a new Item each time the closed tag is named item AND a MyResource is being created (i.e. you have an instance of MyResource somewhere).

When the closed tag is MyResource, add it to a list of results and clean the temporary variable.

public void endElement(String uri, String localName, String qName) throws SAXException {
    if("MyResource".equals(qName)) {
        resources.add(currentResource);
        currentResource = null;

    } else if(currentResource != null && "Item".equals(qName)) {
        currentResource.addItem(new Item(stringBuilder.toString()));
    }
}

I am assuming that you have a List of Item inside MyResource.

Don't forget to add a method to retrieve the resources after the parse :

List<MyResources> getResources() {
    return resources;
}

can you provide code for My XML.As it only contains `` tags and no tags like id,name etc...i.e `` tag name is repeated which is not the case in Employee example i have mentioned...i am totally new to SAX... — Altair, Jul 04 '13 at 04:12

New to SAX Parser so need this BASIC

1 Answers1