[Fixed]-Dynamically add fields to model (or store some fields in a separate class)

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As far as I know, Dynamically injecting fields to a django model is not an easy task. If you want to try , this can be a starting point for you.

Similar post can be found here.
http://www.sixpearls.com/blog/2013/feb/django-class_prepared-signal/

from django.db.models.signals import class_prepared
def add_image_fields(sender, **kwargs):
    """
    class_prepared signal handler that checks for the model massmedia.Image
    and adds sized image fields
    """
    if sender.__name__ == "Image" and sender._meta.app_label == 'massmedia':
        large = models.ImageField(upload_to=".", blank=True, verbose_name=_('large image file'))
        medium = models.ImageField(upload_to=".", blank=True, verbose_name=_('medium image file'))
        small = models.ImageField(upload_to=".", blank=True, verbose_name=_('small image file'))

        large.contribute_to_class(sender, "large")
        medium.contribute_to_class(sender, "medium")
        small.contribute_to_class(sender, "small")

class_prepared.connect(add_image_fields)

which basically attach a function to create those fields.

You can catch class_prepared signal and list all fields, if its a image field, then contribute to class image_field_thumbnail

To get all the fields belongs to a model , you can use get_fields

👤durdenk

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