[Fixed]-Django multiple lookup **kwargs

by

1👍

✅

You did a mistake in sending kwargs, this is how it works…

Lets create one sample lookup dict..

>>> lookup = {'a': 1, 'b': 2}

Simulating create_or_update_if_diff method…

>>> def create_or_update_if_diff(model, defaults=None, **lookup):
...     print lookup

If you assign lookup=lookup while calling the method….this mean you are creating one key word argument called lookup, ideally you are sending only one key word argument 

>>> create_or_update_if_diff('model', lookup=lookup)
{'lookup': {'a': 1, 'b': 2}}

Since you model doesn’t have the lookup field, you are getting FieldError exception.

If you pass lookup dict like below…you are sending two key word arguments a and b to the method..

>>> create_or_update_if_diff('model', **lookup)
{'a': 1, 'b': 2}

So, the logic here is your lookup dict {'a':1, 'b':2} having two keys a and b will be treated like below..

>>> create_or_update_if_diff('model', a=1, b=2)
{'a': 1, 'b': 2}

Hope this helps you to understand how **kwargs works…

👤gsb-eng

0👍

If you do lookup=lookup as parameter, you would get {'lookup': {<your original dict>}} in the function. I think you should just do in your function create_or_update_if_diff:

lookup_params = lookup['lookup']
instance = model.objects.get(lookup_params)
👤Shang Wang

0👍

You are not passing lookup correctly, here is how it suppose to be:

lookup = {'name': default_product_string, 'supplier': supplier}
defaults = {'payment_method': payment_method_instance}
create_or_update_if_diff(defaults=defaults, model=Product, **lookup)
👤lehins

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