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Media.url returns a empty string


How does Django construct the url returned by FileSystemStorage?Django: could not parse the remainder (media i.e. image)How to get an ImageField URL within a template?Django: MEDIA_URL returns Page Not FoundDjango MEDIA_URL not working in templateHow to link my css, js and image file link in djangodjango 1.10 media images don't showDjango Media url returns 404 NOT FOUNDhow to download a file in djangoDjango- empty string in url is not redirecting to the index page






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








0















Whenever I try to image.url function I get an empty string. I also tried the image.path but didn't help. I am using Django 2.15. and I have a Image field in models. My MEDIA_URL returns /media.



I have done Django the config for setting media root, and media url in settings. py also URL pattern.



When I go to the link foo.com/media/imagefile.png it loads perfectly fine. But when I pass through views functions a dictionary



render (request ,templatepath,object.__dict__) 


and in template type



 img.url


I get a empty string.
where as I want to see the real image, or its url mentioned.
Any help is appreciated.
This is what I have in urls.py



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









share|improve this question






























    0















    Whenever I try to image.url function I get an empty string. I also tried the image.path but didn't help. I am using Django 2.15. and I have a Image field in models. My MEDIA_URL returns /media.



    I have done Django the config for setting media root, and media url in settings. py also URL pattern.



    When I go to the link foo.com/media/imagefile.png it loads perfectly fine. But when I pass through views functions a dictionary



    render (request ,templatepath,object.__dict__) 


    and in template type



     img.url


    I get a empty string.
    where as I want to see the real image, or its url mentioned.
    Any help is appreciated.
    This is what I have in urls.py



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









    share|improve this question


























      0












      0








      0








      Whenever I try to image.url function I get an empty string. I also tried the image.path but didn't help. I am using Django 2.15. and I have a Image field in models. My MEDIA_URL returns /media.



      I have done Django the config for setting media root, and media url in settings. py also URL pattern.



      When I go to the link foo.com/media/imagefile.png it loads perfectly fine. But when I pass through views functions a dictionary



      render (request ,templatepath,object.__dict__) 


      and in template type



       img.url


      I get a empty string.
      where as I want to see the real image, or its url mentioned.
      Any help is appreciated.
      This is what I have in urls.py



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









      share|improve this question














      Whenever I try to image.url function I get an empty string. I also tried the image.path but didn't help. I am using Django 2.15. and I have a Image field in models. My MEDIA_URL returns /media.



      I have done Django the config for setting media root, and media url in settings. py also URL pattern.



      When I go to the link foo.com/media/imagefile.png it loads perfectly fine. But when I pass through views functions a dictionary



      render (request ,templatepath,object.__dict__) 


      and in template type



       img.url


      I get a empty string.
      where as I want to see the real image, or its url mentioned.
      Any help is appreciated.
      This is what I have in urls.py



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






      django django-templates






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Mar 28 at 3:08









      Harsh NagarkarHarsh Nagarkar

      158 bronze badges




      158 bronze badges

























          1 Answer
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          The problem is you are passing the template a dictionary. Dictionary object with key img, has no url function to return object. You have to pass the template the whole object.



          example
          objectpassed =Class.object.all()
          render(request,templatepath, 'somename':objectpassed






          share|improve this answer
























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            1 Answer
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            oldest

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            The problem is you are passing the template a dictionary. Dictionary object with key img, has no url function to return object. You have to pass the template the whole object.



            example
            objectpassed =Class.object.all()
            render(request,templatepath, 'somename':objectpassed






            share|improve this answer





























              0
















              The problem is you are passing the template a dictionary. Dictionary object with key img, has no url function to return object. You have to pass the template the whole object.



              example
              objectpassed =Class.object.all()
              render(request,templatepath, 'somename':objectpassed






              share|improve this answer



























                0














                0










                0









                The problem is you are passing the template a dictionary. Dictionary object with key img, has no url function to return object. You have to pass the template the whole object.



                example
                objectpassed =Class.object.all()
                render(request,templatepath, 'somename':objectpassed






                share|improve this answer













                The problem is you are passing the template a dictionary. Dictionary object with key img, has no url function to return object. You have to pass the template the whole object.



                example
                objectpassed =Class.object.all()
                render(request,templatepath, 'somename':objectpassed







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Mar 28 at 3:26









                Harsh NagarkarHarsh Nagarkar

                158 bronze badges




                158 bronze badges





















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