Django - Creating model fields from list with for loop

Question

I have a long list of 100 labels I need my model to have as fields and to also call in succession to access them in other parts of code. I am going to need to modify them in the future so I would like to be able to do it in one place. Is there a simple way to do this. For example:

labels = ['height', 'weight', 'age']

In models.py

class MyModel(models.Model):
    for label in labels:
        label = models.CharField(max_length=255)

Would the above be equal to :

class MyModel(models.Model):
    height = models.CharField(max_length=255)
    weight = models.CharField(max_length=255)
    age = models.CharField(max_length=255)

what does "to also call in succession to access them in other parts of code" mean? Does every model have 100 labels? How often are the labels changing? — dm03514
– dm03514, Commented Jul 13, 2014 at 18:20
So the one model will have 100 different fields then I want to be able to call all those fields. For example: [MyModel.field for field in labels]. — Red-Tune-84
– Red-Tune-84, Commented Jul 13, 2014 at 19:16
No, that code is not equivalent. Strings (fortunately) don't automatically convert to symbols — Eric
– Eric, Commented Jul 13, 2014 at 19:32

knbk · Accepted Answer · 2014-07-13 22:24:54Z

12

Django models have an add_to_class method which adds a field (or any attribute, really) to a class. The syntax is MyModel.add_to_class(name, value). The resulting code would be:

class MyModel(models.Model):
    pass

for label in labels:
    MyModel.add_to_class(label, models.CharField(max_length=255))

Internally, this will call the contribute_to_class method on the value passed, if that method exists. Static attributes are added to the class as-is, but fields have this method and all subsequent processing will ensue.

answered Jul 13, 2014 at 22:24

knbk

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4 Comments

Red-Tune-84 Over a year ago

When would be the best time to call the for loop, not in the model class? In the models.py file? Won't that cause some problems with migrations?

knbk Over a year ago

@SimpleSyrup Just after the model definition. It won't cause problems with syncdb/migrations, as the model and its fields are fully defined after the class is imported.

Red-Tune-84 Over a year ago

I see. Since python looks at the entire file. Thanks for the tips. This solved my dilema.

Nothing here Over a year ago

Is this considered the most "Pythonic" way to write the code if a Model class contains 20+ attributes?

Eric · Accepted Answer · 2014-07-13 21:17:49Z

4

Using locals() should work here:

class MyModel(models.Model):
    for label in labels:
        locals()[label] = models.CharField(max_length=255)

    del locals()['label']

edited Jul 13, 2014 at 21:17

answered Jul 13, 2014 at 19:33

Eric

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4 Comments

Red-Tune-84 Over a year ago

So this is python. making label a local variable for the class. Right?

Eric Over a year ago

Oops, it'll actually be exposed as a property at MyClass.label. You'll want to add a del locals()['label'] after that loop to prevent that

Red-Tune-84 Over a year ago

I don't understand after the entire loop or in the loop? Wouldn't I want it exposed as that is the behavior that django has by default? Or would it actually try to pull it from the class as opposed to the db?

Eric Over a year ago

Without that delete, MyModel.label == 'age', which is not useful to anyone

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