.net如何使用SMTP发送Email

小编给大家分享一下.net如何使用SMTP发送Email，相信大部分人都还不怎么了解，因此分享这篇文章给大家参考一下，希望大家阅读完这篇文章后大有收获，下面让我们一起去了解一下吧！

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 = "allenyinj@gmail.com"; 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, "allen.yin.jun@gmail.com"); 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);

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