How to save list of objects in django models?

Question

I have a list of objects generated after some function, i want it to be saved in django model i tried query set operation

mylist = [(u'england', u'1', u'cde', u'IMG1', u'CP1'), (u'england', u'1', u'cde', u'IMG2', u'CP1'), (u'england', u'2', u'abc', u'IMG1', u'CP1')] 

class Mymodel(models.Model):
   batch_id = models.AutoField(primary_key=True)
   batch_cola = models.CharField(max_length=100)
   batch_colb = models.CharField(max_length=100)
   batch_colc = models.CharField(max_length=100)
   batch_cold = models.CharField(max_length=100)
   batch_cole = models.CharField(max_length=100)

what are you trying to do.please read docs.djangoproject.com/en/1.6/intro/tutorial01/… — sundar nataraj
– sundar nataraj, Commented Jun 24, 2014 at 13:56
are you trying to loop through the list and create three Mymodel instances using the values inside each tuple? — dannymilsom
– dannymilsom, Commented Jun 24, 2014 at 14:21

kreld · Accepted Answer · 2020-06-02 11:52:35Z

20

You can use the bulk_create method - something like:

Mymodel.objects.bulk_create([Mymodel(**{'batch_cola' : m[0],
                                        'batch_colb' : m[1],
                                        'batch_colc' : m[2],
                                        'batch_cold' : m[3],
                                        'batch_cole' : m[4]})
                            for m in mylist])

Instead of iterating over the objects and calling save() on every single one of them, this method results in a single database hit for all the objects, as opposed to one hit for each object.

edited Jun 2, 2020 at 11:52

answered Jun 24, 2014 at 14:33

kreld

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dannymilsom · Accepted Answer · 2014-06-24 14:39:32Z

2

Creating new instances of a model are well documented - see creating objects

But to explicitly anwser your question - one solution is to loop over the list of tuples, pass the appropriate keyword arguments to the model class and then call save(). The last bit is important, otherwise there will be no database level transaction.

for t in mylist:
    Mymodel(batch_cola=t[0], batch_colb=t[1],
           batch_colc=t[2], batch_cold=t[3],
           batch_cole=t[4]).save()

You can also use the convenience create() function from the models manager - you don't need to call .save() then.

for t in mylist:
    Mymodel.object.create(batch_cola=t[0], batch_colb=t[1],
           batch_colc=t[2], batch_cold=t[3], batch_cole=t[4])

The docs say that the two are equivalent - so its somewhat a matter of preference.

On a side note, there is a naming convention in Python you should follow when creating Classes - "Class names should normally use the CapWords convention." - see PEP-8 guidelines

class MyModel(models.Model):
    ...

edited Jun 24, 2014 at 14:39

answered Jun 24, 2014 at 14:32

dannymilsom

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1 Comment

shashisp Over a year ago

thanks a lot, i used Mymodel.objects.create() it works great!!

Mikko Ohtamaa · Accepted Answer · 2014-06-24 14:00:40Z

0

You can use ForeignKey relationship to have a list of objects "contained" by parent object.

Example:

class Parent(models.Model):
    name = models.TextField()

class Child(models.Model):
    name = models.TextField()
    parent = models.ForeignKey(Parent)

Then:

p = Parent.objects.create(name="p")
c = p.child_set.create(name="c")
c2 = p.child_set.create(name="c2")

p.child_set.count()  # Should be 2

For further information refer to ForeignKey documentation in Django.

answered Jun 24, 2014 at 14:00

Mikko Ohtamaa

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