Question

我正在尝试将我的API代码从Sendgrid v2更新为实际的Sendgrid v3，因此我的代码以前看起来像这样：

public void sendCreatedUserEmail(User user) {
    Email from = new Email(FROM);
    from.setName(EMAIL_NAME);
    String subject = "Hello" + user.getName();
    Email to = new Email(user.getEmail());
    Content content = new Content("text/html", "Something");
    Mail mail = new Mail(from, subject, to, content);
    mail.personalization.get(0).addSubstitution("{name1}", user.getName());
    mail.personalization.get(0).addSubstitution("{name2}", user.getName());
    mail.setTemplateId(USER_TEMPLATE_ID);
    SendGrid sg = new SendGrid(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);
    } catch (IOException ex) {
        logger.error(ex);
    }
}

经过几个小时的研究，我将v3更改为：（为了确保画面清晰，我竭尽所能）

public void sendCreatedUserEmail(User user) {
    Mail mail = new Mail();

    Email from = new Email();
    from.setName(EMAIL_NAME);
    from.setEmail(FROM);
    mail.setFrom(from);

    String subject = "Hello, " + user.getName();
    mail.setSubject(subject);

    Personalization personalization = new Personalization();

    Email to = new Email();
    to.setEmail(user.getEmail());
    to.setName(user.getName());
    personalization.addTo(to);

    personalization.setSubject(subject);

    personalization.addSubstitution("{name2}",user.getName());
    personalization.addSubstitution("{name1}",user.getName());

    mail.addPersonalization(personalization);

    Content content = new Content();
    content.setType("text/html");
    content.setValue("Something");
    mail.addContent(content);

    mail.setTemplateId(NEW_USER_TEMPLATE_ID);

    SendGrid sg = new SendGrid(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) {
        logger.error(ex);
    }
}

我收到以下错误：

错误根-java.io.IOException：请求返回的状态代码为400Body：{“ errors”：[{“ message”：“替换不能与动态模板一起使用”，“ field”：“ personalizations.0.substitutions “，” help“：” http://sendgrid.com/docs/API_Reference/Web_API_v3/Mail/errors.html#message.personalizations.substitutions“}]}

我真的不知道该如何进行！我一直在阅读sendgrid文档，但无法理解。

一些可能有用的细节 -Java8是语言 -依赖 -适用于IDE的IntelliJ

很抱歉可能出现的错误，这是我的第一篇文章，英语不是我的主要语言。谢谢！

Answer 1

Sendgrid API的V3使用动态模板数据代替替代。

尝试使用此方法，而不要使用addSubstitution：

personalization.addDynamicTemplateData("{name2}",user.getName());
personalization.addDynamicTemplateData("{name1}",user.getName());

来源：

https://github.com/sendgrid/sendgrid-java/blob/9bc569cbdb908dba609ed0d9d2691dff319ce155/src/main/java/com/sendgrid/helpers/mail/objects/Personalization.java

https://sendgrid.com/docs/ui/sending-email/how-to-send-an-email-with-dynamic-transactional-templates/

Answer 2

尝试：

personalization.addDynamicTemplateData("name2",user.getName());
personalization.addDynamicTemplateData("name1",user.getName());

Sendgrid v3：“替代不能与动态模板一起使用”

2 个答案: