Django Admin - Upload multiple images to multiple model objects in a single upload

user_1510122 0
Viewed 440

I have a model that has close to 150K entries in the Database.
Today, it's possible to upload an image for each entry separately, when editing that specific entry.

We have a need to upload close to 4k image files, one for each entry, and I would like to achieve it by allowing a multiple files upload, where the image file name will match to a name attribute of the model.

Basic model example:

class Entry(models.Model):
    name = models.CharField(unique=True, max_length=200, db_index=True)
    image = models.ImageField(null=True, blank=True)

Is there a way to create a view that will support one upload button for multiple files (where each file represents a different entry of the model in the DB), and handle the rest in an upload handler/storage backend?

django-uploads django-admin django
1 Answers

Update thousands of records at once is challenging as default.

One issue is the performance on the database side. the best I could come up is using update and overwriting some methods. assume you already have your images in some directory.

from django.db.models import fields
from django.db.models import F
from django.db.models.expressions import Value , CombinedExpression
from django.db.models import QuerySet

class TextValue(Value):
    def as_sql(self, compiler, connection):
        connection.ops.check_expression_support(self)
        return '%s', [self.value]

class Expr(F):
    ADD = '||'  # standard concat row value + value in PostgreSQL

    #overwrite method to support text
    def _combine(self, other, connector, reversed):
        if not hasattr(other, 'resolve_expression'):
            other = TextValue(other, output_field=fields.CharField())

        return CombinedExpression(self, connector, other)


class Entry(models.Model):
    name = models.CharField(unique=True, max_length=200, db_index=True)
    image = models.ImageField(upload_to= 'media/' , null=True, blank=True , default='default.png')

now having this piece code you can bulk update

entries= Entry.objects.all() 
entries.update(**{'image': Expr('name') + '.png'})

the best part of doing it like is the performance. this is the only query that is executed

{'sql': 'UPDATE "entry" SET "image" = ("entry"."name" || \'.png\')', 'time': '0.024'}]

update

doing the same thing as admin and saving one instance of model per entry.

from django.contrib import  admin
from django import forms


class EntryForm(forms.ModelForm):
    def __init__(self, *args, **kwargs):
        super(EntryForm, self).__init__(*args, **kwargs)
        self.fields['image'].widget.attrs.update(
            {'multiple': True, 'accept': 'image/jpg,image/png,image/gif', })


class EntryAdmin(admin.ModelAdmin):
    form = EntryForm

    def save_model(self, request, obj, form, change):
        files = request.FILES.getlist('image')

        # if image exist
        if files:
            for image_field in files:
                try:
                    instance = Entry.objects.get(name=image_field.name[:-4])
                    instance.image = image_field
                    instance.save()
                except Entry.DoesNotExist:
                    pass
        else:
            return super().save_model(request, obj, form, change)



admin.site.register(Entry , EntryAdmin)

you can mix these two part to get batter performance

user_37176580
Related