Andriod-在没有用户交互的情况下发送Gmail

时间:2019-07-18 12:42:50

标签: android gmail-api

我正在Android中创建一个没有意图选择器的邮件发送应用程序。

我尝试了GmailSender Link。没有错误,但是邮件没有发送

如何在没有用户交互的情况下发送Gmail邮件?

1 个答案:

答案 0 :(得分:0)

在清单中添加以下权限

<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />

互联网检查

public boolean isOnline() {
    ConnectivityManager cm =
        (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo netInfo = cm.getActiveNetworkInfo();
    if (netInfo != null && netInfo.isConnectedOrConnecting()) {
        return true;
    }
    return false;
}

通过以下参考代码发送邮件

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

Properties props = new Properties();
props.put("mail.smtp.auth", "true");
props.put("mail.smtp.starttls.enable", "true");
props.put("mail.smtp.host", "smtp.gmail.com");
props.put("mail.smtp.port", "587");

Session session = Session.getInstance(props,
  new javax.mail.Authenticator() {
    protected PasswordAuthentication getPasswordAuthentication() {
        return new PasswordAuthentication(username, password);
    }
  });
    try {
        Message message = new MimeMessage(session);
        message.setFrom(new InternetAddress("from-email@gmail.com"));
        message.setRecipients(Message.RecipientType.TO,
            InternetAddress.parse("to-email@gmail.com"));
        message.setSubject("Testing Subject");
        message.setText("Dear Mail Crawler,"
            + "\n\n No spam to my email, please!");

        MimeBodyPart messageBodyPart = new MimeBodyPart();

        Multipart multipart = new MimeMultipart();

        messageBodyPart = new MimeBodyPart();
        String file = "path of file to be attached";
        String fileName = "attachmentName"
        DataSource source = new FileDataSource(file);
        messageBodyPart.setDataHandler(new DataHandler(source));
        messageBodyPart.setFileName(fileName);
        multipart.addBodyPart(messageBodyPart);

        message.setContent(multipart);

        Transport.send(message);

        System.out.println("Done");

    } catch (MessagingException e) {
        throw new RuntimeException(e);
    }