pro*_*eve 8 java jakarta-mail hotmail
我不明白为什么我得到这个例外.这是尝试发送电子邮件的代码.
public void sendAsHotmail() {
final String username = jTextField14.getText();
final String password = jPasswordField4.getText();
String subject = jTextField16.getText();
String Cc = jTextField17.getText();
String Bcc = jTextField18.getText();
String recipient = jTextArea5.getText();
Properties props = new Properties();
props.put( "mail.smtp.host" , "smtp.live.com");
props.put( "mail.smtp.user" , username );
// Use TLS
props.put("mail.smtp.auth" , "true" );
props.put( "mail.smtp.starttls.enable" , "true" );
props.put( "mail.smtp.password" , password );
Session session = Session.getDefaultInstance( props , new Authenticator() {
@Override
protected PasswordAuthentication getPasswordAuthentication() {
if( username == null | password == null )
JOptionPane.showMessageDialog( new JFrame() , "username or password incorrect");
return new PasswordAuthentication( username , password );
}
});
String to = recipient;
String from = username + "@hotmail.com";
String emailMessage = jTextArea2.getText();
MimeMessage message = new MimeMessage(session);
MimeBodyPart attachment = new MimeBodyPart();
MimeBodyPart messagePart = new MimeBodyPart();
FileDataSource fds = new FileDataSource( fileName );
try {
message.setRecipients( Message.RecipientType.TO, InternetAddress.parse( to ) );
message.setFrom( new InternetAddress(from) );
message.setSubject(subject);
message.setText( emailMessage );
attachment.setDataHandler( new DataHandler( fds ) );
attachment.setFileName( fileName );
messagePart.setText( emailMessage );
Multipart hotmailMP = new MimeMultipart();
hotmailMP.addBodyPart(attachment);
hotmailMP.addBodyPart( messagePart );
message.setContent( hotmailMP );
Transport transport = session.getTransport("smtp");
transport.send(message);
JOptionPane.showMessageDialog(new JFrame() , "mail sent !");
} catch(Exception exc) {
System.out.println(exc);
}
}
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为什么我会得到这个例外?如果代码有任何问题,请说明问题所在.
我同意@米美。在您的用户名中,您似乎使用了不完整的用户名(这是Authentication失败的原因)。对于Hotmail,您必须输入Windows Live ID(可能是)xyz@hotmail.com , qrs@gmail.com 。
因此,请输入正确的用户名。并@hotmail.com从from变量中删除。其余代码很好。
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