java SendGrid：如何为发件人添加一个简单的非邮件地址名

Question

我正在使用SendGrid API v3 for Java。它起作用了。然而，如果发送方是，比如说，hello@world.org，那么接收方只会看到非常hello@world.org。我试图做到的是，收件人也会看到一个简单的名称（例如，Hello World <hello@world.org>），如下所示：

（上面，请注意，实际地址是noreply@k...，但前面是Kela Fpa。）

我如何通过编程实现这一点

Answer 1

如果没有您的代码，很难确切地建议该做什么，但根据他们的API文档，端点实际上支持发送者的可选“name”属性

再看一眼他们的Java API源代码，它看起来像这样的示例：

import com.sendgrid.*;
import java.io.IOException;

public class Example {
  public static void main(String[] args) throws IOException {
    Email from = new Email("test@example.com");
    String subject = "Sending with SendGrid is Fun";
    Email to = new Email("test@example.com");
    Content content = new Content("text/plain", "and easy to do anywhere, even with Java");
    Mail mail = new Mail(from, subject, to, content);

    SendGrid sg = new SendGrid(System.getenv("SENDGRID_API_KEY"));
    Request request = new Request();
    try {
      request.setMethod(Method.POST);
      request.setEndpoint("mail/send");
      request.setBody(mail.build());
      Response response = sg.api(request);
      System.out.println(response.getStatusCode());
      System.out.println(response.getBody());
      System.out.println(response.getHeaders());
    } catch (IOException ex) {
      throw ex;
    }
  }
}

使用源代码，您可以向电子邮件构造函数提供一个“名称”，如here 可以重新考虑：

import com.sendgrid.*;
import java.io.IOException;

public class Example {
  public static void main(String[] args) throws IOException {
    Email from = new Email("test@example.com", "John Doe");
    String subject = "Sending with SendGrid is Fun";
    Email to = new Email("test@example.com", "Jane Smith");
    Content content = new Content("text/plain", "and easy to do anywhere, even with Java");
    Mail mail = new Mail(from, subject, to, content);

    SendGrid sg = new SendGrid(System.getenv("SENDGRID_API_KEY"));
    Request request = new Request();
    try {
      request.setMethod(Method.POST);
      request.setEndpoint("mail/send");
      request.setBody(mail.build());
      Response response = sg.api(request);
      System.out.println(response.getStatusCode());
      System.out.println(response.getBody());
      System.out.println(response.getHeaders());
    } catch (IOException ex) {
      throw ex;
    }
  }
}

注意，电子邮件构造函数正在更改

如果您出于某种原因没有使用Mail helper类，请告诉我，我可能可以重新编写一个示例