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Javascript Sendgrid状态202 node.js_Javascript_Java_Node.js_Sendgrid - Fatal编程技术网

Javascript Sendgrid状态202 node.js

Javascript Sendgrid状态202 node.js,javascript,java,node.js,sendgrid,Javascript,Java,Node.js,Sendgrid,使用SendGrid时,我收到以下错误。即使我收到一个错误,邮件仍然被发送 null{statusCode:202,主体:“”,头:{server:'nginx', 日期:2017年7月5日星期三18:50:52 GMT, “内容类型”:“文本/普通;字符集=utf-8”, “内容长度”:“0”, 连接:“关闭”, “x-message-id”:“EUkSULsjRku6bJTkTFsQ9w”, “x帧选项”:“拒绝”, “访问控制允许来源”:“, “访问控制允许方法”:“POST”, “访问控

使用SendGrid时,我收到以下错误。即使我收到一个错误,邮件仍然被发送

null{statusCode:202,主体:“”,头:{server:'nginx', 日期:2017年7月5日星期三18:50:52 GMT, “内容类型”:“文本/普通;字符集=utf-8”, “内容长度”:“0”, 连接:“关闭”, “x-message-id”:“EUkSULsjRku6bJTkTFsQ9w”, “x帧选项”:“拒绝”, “访问控制允许来源”:“, “访问控制允许方法”:“POST”, “访问控制允许标头”:“授权,内容类型,代表,x-sg-elas-acl”, “访问控制最大年龄”:“600”, 'x-no-cors-reason':''} _http_outgoing.js:357 抛出新错误('发送头后无法设置头')

使用以下代码:

var helper = require('sendgrid').mail,
  sg = require('sendgrid')('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');

  var from_email = new helper.Email('my.mail');
  var to_email = new helper.Email('my.mai');
  var subject = 'Subject';
  var content = new helper.Content('text/plain', 'test');
  var mail = new helper.Mail(from_email, subject, to_email, content);

  var request = sg.emptyRequest({
      method: 'POST',
      path: '/v3/mail/send',
      body: mail.toJSON()
  });

  sg.API(request, function(err, response) {
      console.log(err, response);
      if (!err) {
          res.send({
              message: 'An email has been sent to the provided email with further instructions.'
          });
      } else {
          return res.status(400).send({
              message: 'Failure sending email'
          });
      }
  });

我怎样才能解决这个问题

A 202是一条成功消息,而不是错误。这仅仅意味着你的信息被“接受”

您的代码不应该在任何20x响应上检测到任何错误