Kar*_*ran 5 php email sendgrid
我在 Laravel 中集成了 sendgrid 并且我设法在电子邮件中发送了 sendgrid 的电子邮件模板，但我无法替换电子邮件模板中的内容。我正在使用 Sendgrid Web API V3。
我按照下面链接中给出的步骤操作，但它没有用我的动态数据替换模板中的变量。
链接：如何将动态数据传递到在 sendgrid webapp 上设计的电子邮件模板？:-| 发送网格
这是代码
$sg = new \SendGrid('API_KEY');
$request_body = json_decode('{
"personalizations":[
{
"to":[
{
"email":"example@example.com"
}
],
"subject":"Hello World from the SendGrid PHP Library!"
}
],
"from":{
"email":"from@example.com"
},
"content":[
{
"type":"text/html",
"value":"<html><body> -name- </body></html>"
}
],
"sub": {
"-name-": ["Alice"]
},
"template_id":"xxxxxx-xxx-xxxxxxxx"
}');
$mailresponse = $sg->client->mail()->send()->post($request_body);
echo $mailresponse->statusCode();
echo $mailresponse->body();
echo $mailresponse->headers();
请帮忙。
这非常有效，并且比已经发布的解决方案简单得多：
$email = new \SendGrid\Mail\Mail();
$email->setFrom( "from@example.com", "Some guy" );
$email->addTo( "to@example.com", "Another guy" );
$email->setTemplateId(
new \SendGrid\Mail\TemplateId( TEMPLATE_ID )
);
// === Here comes the dynamic template data! ===
$email->addDynamicTemplateDatas( [
'variable1' => 'Some stuff',
'templatesRock' => 'They sure do!'
] );
$sendgrid = new \SendGrid( API_KEY );
$response = $sendgrid->send( $email );
我通过使用另一种方法克服了这个问题。下面是运行良好的代码。可能会帮助某人..
//create mail object
$mail = new \SendGrid\Mail();
//set from
$from = new \SendGrid\Email("SENDER NAME", "SENDER EMAIL");
$mail->setFrom($from);
//set personalization
$personalization = new \SendGrid\Personalization();
$to = new \SendGrid\Email("RECEIVER NAME", "RECEIVER EMAIL");
$personalization->addTo($to);
$personalization->setSubject("SUBJECT");
//add substitutions (Dynamic value to be change in template)
$personalization->addSubstitution(':name', "Any");
$mail->addPersonalization($personalization);
$mail->setTemplateId("TEMPLATE_ID");
//send email
$sg = new \SendGrid("API_KEY");
$response = $sg->client->mail()->send()->post($mail);