cleanest way to glue generated Flask app code (Swagger-Codegen) to backend implementation

some bits flipped picture some bits flipped · Aug 14, 2017 · Viewed 7.2k times · Source

I have:

  1. a library that does [Stuff]
  2. a swagger API definition, which is roughly #1 with minor differences to map cleanly to a REST service
  3. a flask app generated #2 using Swagger-Codegen - eg results in python controller functions roughly one-to-one with #1.

My intent is that the flask app (all generated code) should only handle mapping that actual REST api and parameter parsing to match the API spec coded in swagger. After any parameter parsing (again, generated code) it should call directly over to my (non-generated) backend.

My question is, how best to hook these up withOUT hand-editing the generated python/flask code? (Feedback on my design, or details of a formal design pattern that accomplishes this would be great too; I'm new to this space).

Fresh from the generator, I end up with python functions like:

def create_task(myTaskDefinition):
    """
    comment as specified in swagger.json
    :param myTaskDefinition: json blah blah blah
    :type myTaskDefinition: dict | bytes
    :rtype: ApiResponse
    """
    if connexion.request.is_json:
        myTaskDefinition = MyTaskTypeFromSwagger.from_dict(connexion.request.get_json())
    return 'do some magic!' # swagger codegen inserts this string :)

On the backend I have my actual logic:

def create_task_backend(myTaskDefinition):
    # hand-coded, checked into git: do all the things
    return APIResponse(...)

What is the right way to get create_task() to call create_task_backend()?

Of course if I make breaking changes to my swagger spec I will have to hand-update the non-generated code regardless; however there are many reasons I may want to re-generate my API (say, add/refine the MyTaskTypeFromSwagger class, or skip checking into git the generated code at all) and if I have to hand-edit the generated API code, then all those edits are blown away with each re-generation.

Of course I could script this with a ~simple grammar in eg. pyparsing; but while this is my first time with this issue, it seems likely it's been widely solved already!

Answer

Dudi picture Dudi · Nov 29, 2017

The following approach worked for me:

  • created three directories:

    • src - for my code,
    • src-gen for the swagger generated code,
    • codegen in which I have put a script that generate the server along with a few tricks.
  • I copied all the templates (available in the swagger build) to codegen/templates and edited the controller.mustache to refer to src/server_impl, so it can use my own code. The editing uses the template language so it is generic. Still it is not perfect (I would change a few naming conventions) but it does the job. So, first add to controller.mustache:

from {{packageName}}.server_impl.controllers_impl import {{classname}}_impl

then add instead of return 'do some magic!' the following:

return {{classname}}_impl.{{operationId}}({{#allParams}}{{paramName}}{{^required}}=None{{/required}}{{#hasMore}}, {{/hasMore}}{{/allParams}})
  • Script:
    • The src has a server_impl directory.
    • It creates a symobolic link so that server_impl can be imported as a python module
cd ../src-gen/swagger_server/
ln -s ../../src/server_impl/
cd ../../codegen
java -jar swagger-codegen-cli.jar generate  \
-i /path_to_your_swagger definition.yaml \
-l python-flask \
-o ../src-gen \
-t ./templates
cd ../src-gen/
python3 -m swagger_server