Get email ID

Question

HI,
    How to get the email id of the currently logged in user, using webdynpro for java 2004.
How to send an email using webdynpro for java with attachments.

Answer 1

HI,
Use the following code for sending mail from current login user.
try{
IWDClientUser user = WDClientUser.getCurrentUser();
String mailTo = "[email protected]";
String remarkDetail = "Your proposal has been reviewed.";
String subjectDetail = "Request for new proposal ";
String mailSenderFirstName = user.getSAPUser().getFirstName();
String mailSenderLastName = user.getSAPUser().getLastName();
String host = "<put the SMTP host name>";
String mailBody = remarkDetail + "\n xxxxx xxxx xxxx xxxxx."   +"\n\n Thanks & Regards" +"\n" + " " + mailSenderFirstName +" " + mailSenderLastName;
Authenticator   authenticator = null;
// try{
// IWDClientUser user = WDClientUser.getCurrentUser();
String mailFrom  = user.getSAPUser().getEmail();
InternetAddress [ ]  addresses_from = { new InternetAddress (mailFrom)};
InternetAddress [ ]  addresses_to = {new InternetAddress ( mailTo )};
Properties  properties = new Properties ( );
properties.put ( "mail.smtp.host", host );
MimeMessage  msg = new MimeMessage (
Session.getInstance ( properties, authenticator ) );
msg.addFrom(addresses_from);
msg.setRecipients ( Message.RecipientType.TO , mailTo  );
msg.setRecipients ( Message.RecipientType.CC , addresses_cc  );
//msg.setRecipients ( Message.RecipientType.BCC, addresses_bcc );
msg.setSubject (subjectDetail);
msg.setText ( mailBody );
Transport.send ( msg );
  } catch (AddressException ae)
ae.printStackTrace();
catch (MessagingException me)
me.printStackTrace();
catch(Exception e)
Now Import the followings
import java.util.Properties;
import javax.mail.Authenticator;
import javax.mail.Message;
import javax.mail.MessagingException;
import javax.mail.Session;
import javax.mail.Transport;
import javax.mail.internet.AddressException;
import javax.mail.internet.InternetAddress;
import javax.mail.internet.MimeMessage;
Thanks and regards
Chandan