The starter app has a 'message.py' controller that has functions for 
creating, receiving, and replying to messages (in addition to listing, 
viewing).

The starter app code uses mailgun but it's very similar to sendgrid.  You 
need to setup a webhook in your sendgrid or mailgun account to post the 
inbound message to your app and then the controller function 
(receive_mailgun_message in the starter app) adds it to the message table.

On Thursday, April 21, 2016 at 2:09:21 PM UTC-4, [email protected] wrote:
>
> Thank you guys. Michel, 
>
> message.set_from('myweb2pyapp.com')
>
> If someone response to the email, how do I retrieve it? I need to log into 
> my sandgrid account I am guessing. Not sure why sessage.set_form is my app 
> name.com
>
>
>
>
> On Thursday, April 21, 2016 at 2:00:00 PM UTC-4, Michael Beller wrote:
>>
>> You can use the sendgrid smtp server or their api.
>>
>> https://sendgrid.com/blog/which-protocol-should-i-use-to-send-email-smtp-or-rest/
>>
>> Here is my logic for using their api (I also use mailgun with I like 
>> also):
>>
>> def send_email_via_sendgrid(email_to, email_subject, email_body):
>>     # https://github.com/sendgrid/sendgrid-python
>>
>>     import sendgrid
>>     from sendgrid import SendGridError, SendGridClientError, 
>> SendGridServerError
>>
>>     sg = sendgrid.SendGridClient('<your sendgrid account>', '<your 
>> sendgrid pwd>', raise_errors=True)
>>
>>     message = sendgrid.Mail()
>>     message.add_to(email_to)
>>     message.set_subject(email_subject)
>>     message.set_html(email_body)
>>     #message.set_text('Body')
>>     message.set_from('Your Name <[email protected]>')
>>     #status, msg = sg.send(message)
>>
>>     try:
>>         sg.send(message)
>>     # raise SendGridClientError
>>     except SendGridClientError:
>>         pass
>>     except SendGridServerError:
>>         pass
>>     finally:
>>         return_code = response.status
>>         return locals()
>>
>>
>> On Wednesday, April 13, 2016 at 12:48:13 PM UTC-4, Krishna Bavandlapally 
>> wrote:
>>>
>>> Thank you Leonel Câmara.
>>>
>>> It worked.
>>>
>>> mail.settings.server = 'smtp.sendgrid.net:587'
>>> mail.settings.sender = [email protected]
>>> mail.settings.login = 'sendgrid_username:sendgrid_password'
>>> mail.settings.tls = True
>>>
>>>
>>>
>>> On Wednesday, 13 April 2016 14:29:16 UTC+5:30, Krishna Bavandlapally 
>>> wrote:
>>>>
>>>> How to integrate (default mailing) SendGrid in Web2py ?
>>>>
>>>

-- 
Resources:
- http://web2py.com
- http://web2py.com/book (Documentation)
- http://github.com/web2py/web2py (Source code)
- https://code.google.com/p/web2py/issues/list (Report Issues)
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