How to get value of a particular xml tag having same names and no attribute while parsing xml in sax parser

Question

Actually i was trying to get value of only one particular node but the node tag is at many places in the xml document. here is xml snippet

<t:Address xmlns:t="http://schemas.microsoft.com/exchange/services/2006/types"><t:Name>XYZ Conference Rooms</t:Name><t:EmailAddress>XYZConferenceRooms@abc.co.in</t:EmailAddress><t:RoutingType>SMTP</t:RoutingType><t:MailboxType>PublicDL</t:MailboxType></t:Address><t:Address xmlns:t="http://schemas.microsoft.com/exchange/services/2006/types"><t:Name>ABC Conference Rooms</t:Name><t:EmailAddress>ABCConferenceRooms@abc.co.in</t:EmailAddress><t:RoutingType>SMTP</t:RoutingType><t:MailboxType>PublicDL</t:MailboxType></t:Address>

The thing is i only want value (ABCConferenceRooms@abc.co.in) from the above xml. The above is just a code snippet we have 10 values of email address in xml.

While i am doing this in my parser

    String value;
String url;
String confName;
String confEmail;

@Override
public void startElement(String uri, String localName, String qName,
        Attributes attributes) throws SAXException {

    value = qName;
     confEmail = qName;
     confName = qName;
}

@Override
public void characters(char[] ch, int start, int length)
        throws SAXException {
    if (value.equals("Value")) {
        String OperationName1Text = new String(ch, start, length);
        info.setValue(OperationName1Text);
        Log.d("URL", OperationName1Text);
    }
     else if (confName.equals("t:EmailAddress")) {
     String getEmail = new String(ch, start, length);
     info.setEmail(getEmail);
     Log.d("Email", getEmail);
     }
}

@Override
public void endElement(String uri, String localName, String qName)
        throws SAXException {
    value = qName;
    confEmail = qName;
    confName = qName;
}

i am only able to display all the email address but i only want one mentioned above. What changes in the code is needed.?

Anyone worked on similar situation can help Thanks Mike

Answer 1

当您在两个地址节点中使用相同的名称空间时，它们也将继承到电子邮件节点。因此名称空间没有区别，您找不到确切的节点来获得所需的名称。如果您知道电子邮件节点的位置（第二个电子邮件节点等），那么您可以在startElement（）中拥有一个计数器，当计数器找到第二个电子邮件节点时，便会在其中获取相关文本。

Answer 2

Did you try SAX XML Parser for your work. If not then try it once ...

your XML file is:

<?xml version="1.0"?>
<company>
    <staff>
        <firstname>yong</firstname>
        <lastname>mook kim</lastname>
        <nickname>mkyong</nickname>
        <salary>100000</salary>
    </staff>
    <staff>
        <firstname>low</firstname>
        <lastname>yin fong</lastname>
        <nickname>fong fong</nickname>
        <salary>200000</salary>
    </staff>
</company>

JAVA CODE for XML parser

import javax.xml.parsers.SAXParser;
import javax.xml.parsers.SAXParserFactory;
import org.xml.sax.Attributes;
import org.xml.sax.SAXException;
import org.xml.sax.helpers.DefaultHandler;

public class ReadXMLFile {

   public static void main(String argv[]) {

    try {

    SAXParserFactory factory = SAXParserFactory.newInstance();
    SAXParser saxParser = factory.newSAXParser();

    DefaultHandler handler = new DefaultHandler() {

    boolean bfname = false;
    boolean blname = false;
    boolean bnname = false;
    boolean bsalary = false;

    public void startElement(String uri, String localName,String qName, 
                Attributes attributes) throws SAXException {

        System.out.println("Start Element :" + qName);

        if (qName.equalsIgnoreCase("FIRSTNAME")) {
            bfname = true;
        }

        if (qName.equalsIgnoreCase("LASTNAME")) {
            blname = true;
        }

        if (qName.equalsIgnoreCase("NICKNAME")) {
            bnname = true;
        }

        if (qName.equalsIgnoreCase("SALARY")) {
            bsalary = true;
        }

    }

    public void endElement(String uri, String localName,
        String qName) throws SAXException {

        System.out.println("End Element :" + qName);

    }

    public void characters(char ch[], int start, int length) throws SAXException {

        if (bfname) {
            System.out.println("First Name : " + new String(ch, start, length));
            bfname = false;
        }

        if (blname) {
            System.out.println("Last Name : " + new String(ch, start, length));
            blname = false;
        }

        if (bnname) {
            System.out.println("Nick Name : " + new String(ch, start, length));
            bnname = false;
        }

        if (bsalary) {
            System.out.println("Salary : " + new String(ch, start, length));
            bsalary = false;
        }

    }

     };

       saxParser.parse("c:\\file.xml", handler);

     } catch (Exception e) {
       e.printStackTrace();
     }

   }

}

your out put is :

Start Element :company
Start Element :staff
Start Element :firstname
First Name : yong
End Element :firstname
Start Element :lastname
Last Name : mook kim
End Element :lastname
Start Element :nickname
Nick Name : mkyong
End Element :nickname
Start Element :salary
Salary : 100000
End Element :salary
End Element :staff
Start Element :staff
Start Element :firstname
First Name : low
End Element :firstname
Start Element :lastname
Last Name : yin fong
End Element :lastname
Start Element :nickname
Nick Name : fong fong
End Element :nickname
Start Element :salary
Salary : 200000
End Element :salary
End Element :staff
End Element :company