Question

我已遵循SendGrid文档来创建动态交易电子邮件。但是由于某种原因，我无法分配变量。他们总是返回空。

const sgMail = require("@sendgrid/mail")
const SENDGRID_API_KEY = "deleted for safety, no worries i fill the right value here"
sgMail.setSubstitutionWrappers("{{", "}}")
sgMail.setApiKey(SENDGRID_API_KEY)

const msg = {
    to: "deleted for safety, no worries i fill the right value here",
    from: "deleted for safety, no worries i fill the right value here",
    subject: "Hello world",
    text: "Hello plain world!",
    html: "<p>Hello HTML world!</p>",
    templateId: "deleted for safety, no worries i fill the right value here",
    substitutions: {
        name: "Some One",
        city: "Denver",
    },
};
sgMail.send(msg);

模板：

<html>
<head>
    <title></title>
</head>
<body>
Hello {{name}},
<br /><br/>
I'm glad you are trying out the template feature!
<br><br>
I hope you are having a great day in {{city}} :)
<br /><br/>
</body>
</html>

结果：

你好，

很高兴您正在尝试使用模板功能！

我希望您在:)过得愉快。

Api密钥正确，结果是我收到的邮件。你们能告诉我我在想什么吗？

Answer 1

安装最新版本的@ sendgrid / mail软件包，然后按照以下官方文档链接 Transactional Templates Use Case

现在，您必须使用dynamic_template_data而不是替换。您也可以删除此行

sgMail.setSubstitutionWrappers（“ {{”，“}}”）

因为从v3 API开始，所以无需指定替换包装器，因为它将假定您正在使用车把。

这是一个应该起作用的示例：

 const sgMail = require('@sendgrid/mail');
   sgMail.setApiKey(process.env.SENDGRID_API_KEY);
   const msg = {
     to: 'recipient@example.org',
     from: 'sender@example.org',
     templateId: 'd-f43daeeaef504760851f727007e0b5d0',
     dynamic_template_data: {
     subject: 'Testing Templates',
        name: 'Some One',
        city: 'Denver',
      },
   };

sgMail.send(msg);

node.js sendgrid实现

1 个答案: