sendgrid-python should define it's own Error class

[w-]

Issue Summary

related to sendgrid/python-http-client#16

Crurently if an application/library consumes sendgrid-python and an API request surfaces a HTTP error, it raises a urllib HTTPError.

Now, in order to handle this specific error, the consuming application needs to import urllib.

Ideally, a consuming application only needs to know about sendgrid-python and not the underlying implementation of python-http-client or urllib in order to handle common place http errors.

Steps to Reproduce

1. make an api call that returns anything other than HTTP2xx
2. attempt to handle error raised without importing urllib and without doing a catch all.

Technical details:

• sendgrid-python Version: master (latest commit: [commit number])
• Python Version: all

[dibyadas]

Hi @w- ! I was working on the same thing with ref to the issue #224. We can define a custom SendGrid exception.
One approach could be to raise a SendGridException before the request is made, for those requests that we know are going to return an error code. For ex :- an email request with empty content or empty subject.
Other approach could be to wrap the existing SendGridAPIClient class in a 'SendGridClient' class that will raise SendGridException based on the response codes, among other things.
What are your views on this?

[w-]

Hi @dibyadas ,

My specific case here is that there is no way to handle an erronous HTTP response because the underlying urllib will always throw an HTTPError.

IMHO, we should have a specific Error class that represents specifically when the API call returns an non-success HTTP Response. I have not given much thought to the handling of other errors such as the example you provide "where an email request with empty content or empty subject". I just think those are a seperate class of errors.

The sendgrid API V3 documentation outlines how the API returns relatively rich error information with the body of the response often outlining specific issues with the API request.
In the current implementation it is not possible to take advantage of this without having to import urllib at the application level.

Given the specific implementation of sendgrid-python and python-http-client I'm not sure what the best way to achieve this would be.

[thinkingserious]

Thanks for the feedback @w-!

@dibyadas has now created a PR. It would be great if you could take a look and let us know your thoughts.

[w-]

@thinkingserious thanks for taking a look at this.
I'm not sure that identifying this as "duplicate" and closing this is appropriate. I feel that this and #224 are completely different topics.
Especially after reviewing the changes from @dibyadas it is clear his focus is around sending mail, whereas mine is around use of the API library in general.
Again, currently any API call made using sendgrid-api that doesn't result in an http2xx returns a urllib HTTPError. There is no way for a consuming application to handle these errors specifically without understanding that sendgrid uses http-python-client and that uses urllib.

[thinkingserious]

Please see the comment I made here.

Thanks for taking the time to help out with your feedback, we greatly appreciate it!

[phrohdoh]

I'd like to resurrect this issue as sendgrid-python the package still does not export its own error types.

This means, as @w- pointed out, library users must be knowledgeable of an implementation detail (urllib, python-http-client, etc).

Due to this library users must declare python-http-client as an explicit dependency (which means keeping python-http-client and sendgrid in sync).

IMO ideally sendgrid would re-export the error types defined in python-http-client which would allow sendgrid library users to write code similar to the following:

import sendgrid
from sendgrid.helpers.mail import *

sg = sendgrid.SendGridAPIClient(apikey='foobar')
from_email = Email('foo@bar.com', name='Foo')
to_email = Email('qux@baz.net', name = 'Qux')
subject = 'my subject'
content = Content('text/html', 'some html here')
mail = Mail(from_email, subject, to_email, content)

try:
    response = sg.client.mail.send.post(request_body=mail.get())
except sendgrid.exceptions.HTTPError as e:
    print(e.read())