Django rest framework api - image url is not returning properly

Question

I am facing problem in returned image url, which is not proper.

My return image url is "http://127.0.0.1:8000/showimage/6/E%3A/workspace/tutorial_2/media/Capture1.PNG" But i need

"http://127.0.0.1:8000/media/Capture1.PNG"

When i click on image_url then image open in new browser tab But currently its shown error: view.py

from showimage.models import ShowImage
from showimage.serializers import ShowImageSerializer
from rest_framework import generics

# Create your views here.

    class ShowImageList(generics.ListCreateAPIView):
        queryset = ShowImage.objects.all()
        serializer_class = ShowImageSerializer
        
    class ShowImageDetail(generics.RetrieveUpdateDestroyAPIView):
        queryset = ShowImage.objects.all()
        serializer_class = ShowImageSerializer

model.py

from __future__ import unicode_literals

from django.db import models
from django.conf import settings


# Create your models here.

class ShowImage(models.Model):
    image_name = models.CharField(max_length=255)
    image_url = models.ImageField(upload_to=settings.MEDIA)

serializer.py

from rest_framework import serializers
from showimage.models import ShowImage

class ShowImageSerializer (serializers.ModelSerializer):
    class Meta:
        model = ShowImage
        fields = ('id', 'image_name', 'image_url')

settings.py

MEDIA=os.path.join(BASE_DIR, "media")

urls.py

from django.conf.urls import url, include
from django.contrib import admin

urlpatterns = [
    url(r'^admin/', admin.site.urls),
    url(r'^showimage/', include('showimage.urls')),
]

I am new in python and also in django-rest-framework. Please also tell me how we extend models or serialize class

What do you mean by "extend models". What do you want to achieve? — Remi Smirra
– Remi Smirra, Commented Feb 18, 2016 at 8:33
All you're image urls look very strange and wrong... the one in your error image (which you should include as text btw) is trying to link to an E drive which is useless to everyone involved in that request. and then your others shouldn't care what port or domain its hosted on. — Sayse
– Sayse, Commented Feb 18, 2016 at 8:38

Remi Smirra · Accepted Answer · 2016-02-18 08:33:08Z

9

You might want to try this in your settings:

MEDIA_URL = '/media/'
MEDIA_ROOT=os.path.join(BASE_DIR, "media")

urlpatterns = [
    url(r'^admin/', admin.site.urls),
    url(r'^showimage/', include('showimage.urls')),
]

urlpatterns += static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

And in your models:

class ShowImage(models.Model):
    image_name = models.CharField(max_length=255)
    image_url = models.ImageField(upload_to="") # or upload_to="images", which would result in your images being at "http://127.0.0.1:8000/media/images/Capture1.PNG"

answered Feb 18, 2016 at 8:33

Remi Smirra

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SuReSh · Accepted Answer · 2018-04-26 03:30:00Z

7

Finally, i solve this road block with the help of @Remi Thanks @Remi But some other change i do so that i elaborate solution and fix this issue.

settings.py

STATIC_URL = '/static/'
MEDIA_URL = '/media/'
MEDIA_ROOT=os.path.join(BASE_DIR, "media")

urls.py

from django.conf.urls import url, include
from django.contrib import admin
from django.conf import settings
from django.conf.urls.static import static

urlpatterns = [
    url(r'^admin/', admin.site.urls),
    url(r'^showimage/', include('showimage.urls')),
]

urlpatterns += static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

edited Apr 26, 2018 at 3:30

answered Feb 18, 2016 at 10:08

SuReSh

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Comments

Kamlesh · Accepted Answer · 2016-02-18 08:35:27Z

1

Your code seems correct except one thing you have passed settings.MEDIA in uploads image. you don't need to pass settings.MEDIA in uploads.

try this

image_url = models.ImageField(upload_to='Dir_name')

Dir_name will create when you'll run script.

answered Feb 18, 2016 at 8:35

Kamlesh

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