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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;
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
add a comment |
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
add a comment |
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
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
django django-templates
asked Mar 28 at 3:08
Harsh NagarkarHarsh Nagarkar
158 bronze badges
158 bronze badges
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
add a comment |
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
add a comment |
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
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
answered Mar 28 at 3:26
Harsh NagarkarHarsh Nagarkar
158 bronze badges
158 bronze badges
add a comment |
add a comment |
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