generate a Django-REST-API directly ?

Question

I have already a database that contains 100 tables.

Now, I would like to expose it as a REST-API.

I've done this before so I am familiar with the procedure. I create a serializer for the model, a ViewSet for the Serialiizer and an a router. I test it on 2 tables and it works.

The problem is that I want to expose all my 100 tables. And doing the same prcedure 100 times is no fun.

Is there a way to do it only one time?

Answer 1

It appears there is no way to do so in django.

NOW, THIS IS NOT THE BEST SOLUTION,

SOMETIMES, DIFFERENT SERIALIZERS CLASSES ARE NEEDED DEPENDING ON RELATIONS

I just created a script that works:

You just need to run it from your project directory with your app name as an argument

`import os
import sys

ROUTER_TEMPLATE = "router.register(r'{}', views.{}ViewSet)"
SERIALIZER_TEMPLATE = """
class {}Serializer(serializers.HyperlinkedModelSerializer):

    class Meta:
        model = {}
        fields = '__all__'
"""
VIEW_TEMPLATE= """

class {}ViewSet(viewsets.ModelViewSet):
    queryset = {}.objects.all()
    serializer_class = {}Serializer


"""

def gen_api(app_name):
        models= open(app_name+'/models.py', 'r')
    aLines = models.readlines()

    for sLine in aLines:

        aLine = sLine.split()

        for sWord in aLine:

            if sWord == "class" and aLine[aLine.index(sWord)+1] != "Meta:":
                model_name= aLine[aLine.index(sWord)+1].split("(")[0]
                gen(app_name,model_name )




def gen(app_name, name):

    config = open(app_name+'/urls.py', 'a') #append
    config.write("\n"+ ROUTER_TEMPLATE.format(name, name))
    config.close()

    config = open(app_name+'/serializers.py', 'a') #append
    config.write("\n"+ SERIALIZER_TEMPLATE.format(name, name))
    config.close()

    config = open(app_name+'/views.py', 'a') #append
    config.write("\n"+ VIEW_TEMPLATE.format(name, name, name))
    config.close()

if __name__ == "__main__":

    gen_api(sys.argv[1])
`