'Sendgrid setup on SQL Database Mail
I have been trying to setup sendgrid on SQL Database Mail but all the time it says Cannot send mail to mail server.
This is what i have done.
- I have sendgrid accound with [email protected] and created api key and SG.xxxx password.
- As this is prod setup instance, i had done domain level authentication and my webmaster updated CNAME as requested by sendgrid.
- Tested using TELNET smtp.sendgrid.net 587 and authenticated with SG.xxxx password Base64 encoded and this was able to send email from [email protected] to [email protected]
- I setup profile with [email protected] as SQL notification Profile and selected Basic Authentication and provided 'apikey' as username and 'SG.xxx' as password in Database Mail - 'Configure Database Mail' option in SQL SSMS. Unchecked - 'The server requires secure authentication (SSL) connection' Port 587
- I tried to 'Send Test Mail' with '[email protected]' to '[email protected]' but the Database Mail reflects as below screenshot error
I have tried using [email protected] and sendgrid account password in Database Mail Config wizard for Basic authentication, but sendgrid suggested they don't accept Basic authentication. Please help how and where should i pass the api key if not under Basic Authentication ? I tried updating with msdb.dbo.sysmail_add_account_sp nad updating records but it too didn't resolve and send email from Database mail.
Any help/ guidance would be greatly appreciated or if you had faced this how have you resolved it.
Solution 1:[1]
Hi the solution is as follows
Create a dbmail profile
- server name: smtp.sendgrid.net
- port : 25
SMTP Authentication
- Basic Auth
- User name: apikey
- Password: : {the api-key for your sendgrid account}
Thats it. Send a test email.
If it still doesn't work, if you are on AWS EC2 ensure that security group allows outbound port:25
Here is the link from sendgrid about apikey authentication. https://docs.sendgrid.com/for-developers/sending-email/upgrade-your-authentication-method-to-api-keys
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | jcrawfor74 |

