and this is what i get when i send the request from front-end

Take a look at this maybe it's useful. I think that your problem is related with the permissions, so look this https://www.django-rest-framework.org/api-guide/permissions/

Hello everyone! Could anyone help me please. I need to make a field like this: Class Person(models. Model) : select = ( ('A', ('name_of_category ', ('1', '2', '3'...), 'name_of_category-2', (('1', '2'), ('3', '4')...)...), ('B', (('name_of_category ', ('1', '2', '3'...), 'name_of_category-2', (('1', '2'), ('3', '4')...)...)) ) menu = models.CharField (max_length=100, choices = select) But I couldn't make it by M2MField. Is it possible to make a field like this, without using M2MField? Or is there any example for this type of fields?

Take a look at this maybe it's useful. I think that your problem is related with the permissions, so look this https://www.django-rest-framework.org/api-guide/permissions/

Hey @pyflare you up for some mentoring ?

Mentoring for ?

Im still not sure how I can get multiple pictures done in one create view in the admin app.

Its a huge project, I cant share that here. I could give you access on github ?

Could you share just the snippet related to uploading mutliple images ?

The link I shared i get, you inject a form that lets you upload pictures and then you connect them when you override the save method in admin.py ( I had no idea you could do that )

Ah ok 😂 Anyway I can tell one of the code flow in a classic view when you have Item and ItemImage as models When you upload multiple images , in the request object you can get them via request.FILES.getlist("name_of_input") and loop it to create ItemImage objects

Yeah, i got that now reading the "russian" post. I think what went wrong is I had no idea you could override the ADMIN save method, and so I thought you were talking about he MODEL save method ;)

plz help me

I think I screwed up the return from the upload_to function.

All you need is the model inline, no need to override the save...

appaearently i dont have to do that, this works

Thanks for playing ball Mirco... It helpes to have someone to "play ball" with ;)

Share settings

Mmm you should check official docs for better explanation

It's becouse the fieldname (item) on my inline model (ItemPicture) matches the parent model: https://docs.djangoproject.com/en/2.2/ref/contrib/admin/#django.contrib.admin.TabularInline So I have my self to thank for using good practice and django to thank for makes it super intuitive...

http://dpaste.com/238ZTJF

You have to specify dirs where find templates or Django will look for into your app dir

I threw that in a comment in the model, I will never remember that... :D

Or this most specific https://www.django-rest-framework.org/api-guide/permissions/#how-permissions-are-determined

https://pastebin.com/nBaryszA

you still working on your django admin site?

no, this is the API representation of the pictures after i used multiple pictures.

I have this particular Model say: Relation with 2 column: Parent | Children And here are some example entries: A | B A | P B | C C | D D | E P | Q R | S I am trying to use ModelManager to validate any new entry so no circular relation should happen. Suppose I try to enter new Relation E | A I will call Relation.objects.is_relation_possible(E, A) It should return False as E cannot be Parent of A. Similarly, Q cannot be a parent of A But R and S can be a parent of A or any other. Is there any easy way other than looping over each parent of every immediate parent..

how is your is_relation_possible() looks like?

Thinking something like this: class RelationManager(models.Manager): def relation_not_possible(self, group_id, parent_id, children_id): person = Person.objects.get(pk=parent_id) return person.child_relations.filter(parent_id=parent_id, group_id=group_id).first() class Relation(models.Model): group_id = IntegerField(null=True) parent = ForeignKey(Person, related_name="parent_relations", on_delete=models.PROTECT) children = ForeignKey(Person, related_name="child_relations", on_delete=models.PROTECT) objects = RelationManager() class Meta: unique_together = ('group', 'parent', 'children')

For readability https://pastebin.com/r6QNVi6i

seems like not tho

I have this particular Model say: Relation with 2 column: Parent | Children And here are some example entries: A | B A | P B | C C | D D | E P | Q R | S I am trying to use ModelManager to validate any new entry so no circular relation should happen. Suppose I try to enter new Relation E | A I will call Relation.objects.is_relation_possible(E, A) It should return False as E cannot be Parent of A. Similarly, Q cannot be a parent of A But R and S can be a parent of A or any other. Is there any easy way other than looping over each parent of every immediate parent..

@pyflare https://pastebin.com/nBaryszA Look at the serializer... This is all it took to add the pictures... awsome.

thant is interesting thanks a lot, but the requests are done by the same user who is already a superuser

i cant give you "easier way" if i dont even know what to compare tho

you can, new_comment is just a variable

Hello, it is a possible to create a widget (or live-chat) based on Django? If it is possible, i try to find examples\courses for this problem. Thank for atention!)