創建了 sendgrid email 模板，如何從 Java 調用它

Question

我在 sendgrid 中創建了 email 模板 - 具有可替換的值；

我從 rabbitMQ 隊列中獲取 JSON 有效載荷（包含替代值）以處理 email。 我的問題是如何從 Java 調用 sendgrid email 模板？

Answer 1

我找到了解決方案，通過sendgrid調用sendgrid模板和email的方法如下：

SendGrid sendGrid = new SendGrid("username","password"));
SendGrid.Email email = new SendGrid.Email();

//Fill the required fields of email
email.setTo(new String[] { "xyz_to@gmail.com"});
email.setFrom("xyz_from@gmail.com");
email.setSubject(" ");
email.setText(" ");
email.setHtml(" ");

// Substitute template ID
email.addFilter(
    "templates",
    "template_id",
    "1231_1212_2323_3232");

//place holders in template, dynamically fill values in template
email.addSubstitution(
     ":firstName",
     new String[] { firstName });
email.addSubstitution(
     ":lastName",
     new String[] { lastName });

// send your email
Response response = sendGrid.send(email);

Answer 2

這是我的工作解決方案。

import com.fasterxml.jackson.annotation.JsonProperty;
import com.sendgrid.*;
import java.io.IOException;
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
public class MailUtil {
    public static void main(String[] args) throws IOException {
        Email from = new Email();
        from.setEmail("fromEmail");
        from.setName("Samay");
        String subject = "Sending with SendGrid is Fun";
        Email to = new Email();
        to.setName("Sam");
        to.setEmail("ToEmail");
        DynamicTemplatePersonalization personalization = new DynamicTemplatePersonalization();
        personalization.addTo(to);
        Mail mail = new Mail();
        mail.setFrom(from);
        personalization.addDynamicTemplateData("name", "Sam");
        mail.addPersonalization(personalization);
        mail.setTemplateId("TEMPLATE-ID");
        SendGrid sg = new 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;
        }
    }
    private static class DynamicTemplatePersonalization extends Personalization {

        @JsonProperty(value = "dynamic_template_data")
        private Map<String, String> dynamic_template_data;

        @JsonProperty("dynamic_template_data")
        public Map<String, String> getDynamicTemplateData() {
            if (dynamic_template_data == null) {
                return Collections.<String, String>emptyMap();
            }
            return dynamic_template_data;
        }

        public void addDynamicTemplateData(String key, String value) {
            if (dynamic_template_data == null) {
                dynamic_template_data = new HashMap<String, String>();
                dynamic_template_data.put(key, value);
            } else {
                dynamic_template_data.put(key, value);
            }
        }

    }
} 

Answer 3

這是來自最新 API 規范的示例：

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 = "I'm replacing the subject tag";
    Email to = new Email("test@example.com");
    Content content = new Content("text/html", "I'm replacing the <strong>body tag</strong>");
    Mail mail = new Mail(from, subject, to, content);
    mail.personalization.get(0).addSubstitution("-name-", "Example User");
    mail.personalization.get(0).addSubstitution("-city-", "Denver");
    mail.setTemplateId("13b8f94f-bcae-4ec6-b752-70d6cb59f932");

    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;
    }
  }
}

https://github.com/sendgrid/sendgrid-java/blob/master/USE_CASES.md

Answer 4

最近 Sendgrid 升級了 Maven 版本 v4.3.0 這樣您就不必為動態數據內容創建額外的類。 從此鏈接閱讀更多信息https://github.com/sendgrid/sendgrid-java/pull/449

Answer 5

如果您正在執行spring-boot maven 項目，則需要在 pom.xml 中添加sendgrid-java maven 依賴pom.xml 。 同樣，在項目resource文件夾下的application.properties文件中，分別在spring.sendgrid.api-key=SG.xyz和templateId=d-cabc等屬性下添加SENDGRID API KEY AND TEMPLATE ID 。

完成了預先設置。 您可以創建一個簡單的 controller class，如下所示：

快樂編碼！

@RestController
@RequestMapping("/sendgrid")
public class MailResource {

    private final SendGrid sendGrid;

    @Value("${templateId}")
    private String EMAIL_TEMPLATE_ID;

    public MailResource(SendGrid sendGrid) {
        this.sendGrid = sendGrid;
    }

    @GetMapping("/test")
    public String sendEmailWithSendGrid(@RequestParam("msg") String message) {

        Email from = new Email("bijay.shrestha@f1soft.com");
        String subject = "Welcome Fonesal Unit to SendGrid";
        Email to = new Email("birat.bohora@f1soft.com");
        Content content = new Content("text/html", message);

        Mail mail = new Mail(from, subject, to, content);

        mail.setReplyTo(new Email("bijay.shrestha@f1soft.com"));
        mail.setTemplateId(EMAIL_TEMPLATE_ID);

        Request request = new Request();
        Response response = null;


        try {
            request.setMethod(Method.POST);
            request.setEndpoint("mail/send");
            request.setBody(mail.build());

            response = sendGrid.api(request);

            System.out.println(response.getStatusCode());
            System.out.println(response.getBody());
            System.out.println(response.getHeaders());
        } catch (IOException ex) {
            System.out.println(ex.getMessage());
        }

        return "Email was successfully sent";
    }

}

Answer 6

這是基於com.sendgrid:sendgrid-java:4.8.3

        Email from = new Email(fromEmail);
        Email to = new Email(toEmail);
        String subject = "subject";

        Content content = new Content(TEXT_HTML, "dummy value");

        Mail mail = new Mail(from, subject, to, content);

        // Using template to send Email, so subject and content will be ignored
        mail.setTemplateId(request.getTemplateId());

        SendGrid sendGrid = new SendGrid(SEND_GRID_API_KEY);
        Request sendRequest = new Request();

        sendRequest.setMethod(Method.POST);
        sendRequest.setEndpoint("mail/send");
        sendRequest.setBody(mail.build());

        Response response = sendGrid.api(sendRequest);

您還可以在這里找到完整的示例： 完整的 Email object Java 示例