Question

如何保持java邮件传输对象的活动或连接。

我在我的代码中用Web应用程序中的一个简单类文件编写了这个： -

@Resource(name = "myMailServer")
    private Session mailSession;

    Transport transport ;

public boolean sendMail(String recipient, String subject, String text) {
    boolean exe = false;

    Properties p = new Properties();

    String username = "someone@gmail.com";
    String password = "password";

    InitialContext c = null;

    try
    {
         c = new InitialContext();
          mailSession = (javax.mail.Session) c.lookup("java:comp/env/myMailServer");
    }
    catch(NamingException ne)
    {
        ne.printStackTrace();
    }

    try
    {
        Message msg = new MimeMessage(mailSession);
        msg.setRecipients(Message.RecipientType.TO,InternetAddress.parse(recipient, false));
        msg.setSubject(subject);
        msg.setText(text);
        msg.setHeader("MIME-Version" , "1.0" );
        msg.setHeader("Content-Type" , "text/html" );
        msg.setHeader("X-Mailer", "Recommend-It Mailer V2.03c02");
        msg.saveChanges();

        //Transport.send(msg);
        if(transport == null) {
            transport = mailSession.getTransport("smtps");
            System.out.println("" + transport.isConnected());
            if(!transport.isConnected()) {
                transport.connect(username, password);
            }
        }

        transport.sendMessage(msg, msg.getAllRecipients());
        exe = true;
    }
    catch (AddressException e)
    {
        e.printStackTrace();
        exe = false;
    }
    catch (MessagingException e)
    {
        e.printStackTrace();
        exe = false;
    }
    finally {
        /*try {
            if(transport != null)
                transport.close();
        }
        catch(MessagingException me) {
            me.printStackTrace();
        }
        catch(Exception e) {
            e.printStackTrace();
        }*/
    }

    return exe;
}

完整代码here
现在每次运行此代码都需要一些时间来连接邮件服务器
和行

System.out.println("" + transport.isConnected());

打印错误的

如何保留对象传输，因为它确实为空并进入块

if(transport == null) {

或传输对象保持连接...

谢谢你 Pradyut

Answer 1

代码应该是......
使用传输对象的静态初始化
没有任何问题但可以很好的功能 静态传输getTransport（）方法

@Resource(name = "myMailServer")
    private Session mailSession;

    static Transport transport ;

    public boolean sendMail(String recipient, String subject, String text) {
    boolean exe = false;

    Properties p = new Properties();

    String username = "someone@gmail.com";
    String password = "password";

    InitialContext c = null;

    try
    {
         c = new InitialContext();
          mailSession = (javax.mail.Session) c.lookup("java:comp/env/myMailServer");
    }
    catch(NamingException ne)
    {
        ne.printStackTrace();
    }

    try
    {
        Message msg = new MimeMessage(mailSession);
        msg.setRecipients(Message.RecipientType.TO,InternetAddress.parse(recipient, false));
        msg.setSubject(subject);
        msg.setText(text);
        msg.setHeader("MIME-Version" , "1.0" );
        msg.setHeader("Content-Type" , "text/html" );
        msg.setHeader("X-Mailer", "Recommend-It Mailer V2.03c02");
        msg.saveChanges();

        //Transport.send(msg);
        if(transport == null) {

            transport = mailSession.getTransport("smtps");


        }
        if(!transport.isConnected()) {

                transport.connect(username, password);
            }

        transport.sendMessage(msg, msg.getAllRecipients());
        exe = true;
    }
    catch (AddressException e)
    {
        e.printStackTrace();
        exe = false;
    }
    catch (MessagingException e)
    {
        e.printStackTrace();
        exe = false;
    }
    finally {
        /*try {
            if(transport != null)
                transport.close();
        }
        catch(MessagingException me) {
            me.printStackTrace();
        }
        catch(Exception e) {
            e.printStackTrace();
        }*/
    }

    return exe;
}

谢谢你问候
Pradyut

java邮件保持传输对象已连接

1 个答案: