.net SMTP发送Email实例(可带附件)

复制代码 代码如下:

public static void sendEmail(string toAddress, string emailbody)
{
var fromAddress = ConfigurationManager.AppSettings["EmailAddress"];
string fromPassword = ConfigurationManager.AppSettings["EmailPassword"].ToString();
const string subject = "Job Recommendation";
var smtp = new SmtpClient
{
Host = ConfigurationManager.AppSettings["SmtpServer"].ToString(),
Port = int.Parse(ConfigurationManager.AppSettings["SmtpPort"]),
EnableSsl = true,
DeliveryMethod = SmtpDeliveryMethod.Network,
UseDefaultCredentials = false,
Credentials = new NetworkCredential(fromAddress, fromPassword)
};
using (var message = new MailMessage(fromAddress, toAddress, subject, HttpUtility.HtmlEncode(emailbody)))
{
smtp.Send(message);
}
}
//Email Address
//Emial PWD


<--带附件版本->
var fromAddress = "[email protected]";
string fromPassword = "yj1989120";
const string subject = "CV";
var smtp = new SmtpClient
{
Host = "smtp.gmail.com",
Port = 587,
EnableSsl = true,
DeliveryMethod = SmtpDeliveryMethod.Network,
UseDefaultCredentials = false,
Credentials = new NetworkCredential(fromAddress, fromPassword)
};
MailMessage email=new MailMessage(fromAddress, "[email protected]");
email.Subject = "INLINE attachment TEST";
email.IsBodyHtml = true;
string attachmentPath = "C:\\3.jpeg";
Attachment inline = new Attachment(attachmentPath);
inline.ContentDisposition.Inline = true;
inline.ContentDisposition.DispositionType = DispositionTypeNames.Inline;
//inline.ContentId = "1";
//inline.ContentType.MediaType = "image/png";
inline.ContentType.Name = Path.GetFileName(attachmentPath);
email.Attachments.Add(inline);
email.Body = "test";
smtp.Send(email);
email.Dispose();
//如果没有路径，用Stream
Attachment letter = new Attachment(FileUploadLetter.FileContent, FileUploadLetter.PostedFile.ContentType);
letter.ContentDisposition.Inline = true;
letter.ContentDisposition.DispositionType = DispositionTypeNames.Inline;
//inline.ContentId = "1";
letter.ContentType.MediaType = FileUploadLetter.PostedFile.ContentType;
letter.ContentType.Name = Path.GetFileName(FileUploadLetter.PostedFile.FileName);
letter.Name = Path.GetFileName(FileUploadLetter.PostedFile.FileName);