I'm terribly sorry, my bad. Of course, the second example is:
class Image(models.Model):
image = models.ImageField(upload_to = upload_path('i'), blank=True)
thumbnail = models.ImageField(upload_to = upload_path('t'), blank=True)
суббота, 26 января 2013 г., 3:58:55 UTC+4 п
There is no t_upload_path
2013/1/25 Sammael
> Hello, folks!
>
> Here is the relevant part of my model:
>
> def i_upload_to(instance, filename):
> return '/'.join([strftime('i/%y%m/%d'), filename])
>
> def t_upload_to(instance, filename):
> return '/'.join([strftime('t/%y
Hello, folks!
Here is the relevant part of my model:
def i_upload_to(instance, filename):
return '/'.join([strftime('i/%y%m/%d'), filename])
def t_upload_to(instance, filename):
return '/'.join([strftime('t/%y%m/%d'), filename])
class Image(models.Model):
ima
solved this by moving it to the model clean() method, and using
imagefield.file.open() and imagefield.file.read() to get the data for
computing the other fields.
On Wed, May 11, 2011 at 1:28 PM, Brian Craft wrote:
> I have a model with an imagefield, and some fields that are computed
> from the i
I have a model with an imagefield, and some fields that are computed
from the image. I'm trying to override the "save" method to fill in
the other fields.
When I test this in the admin, there are two problems: first, the path
(mymodel.image.path) doesn't honor the upload_to setting on the field.
I
Is there any way to get the full path of file, which will be uploaded?
in save_model action i can get obj.image but it gives me only the file name.
I want to get path with upload_to from model.
Is it possible?
I need it, because I want to make resizing on my image and make its name
with hash to
Hi Javier,
I am experiencing the same problem with 'upload_to'. Have you found
the problem?
Thx,
elm
On 14 nov, 18:31, Javier <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I've create a model:
>
> class ImagenesLugar(models.Model):
> archivo = models.ImageField(upload_to="imageneslugar/")
> te
On Nov 14, 9:31 pm, Javier <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I've create a model:
>
> class ImagenesLugar(models.Model):
> archivo = models.ImageField(upload_to="imageneslugar/")
> texto = models.CharField(max_length=400)
>
> The thing is that when i create a new instance of this mode
Hi,
I've create a model:
class ImagenesLugar(models.Model):
archivo = models.ImageField(upload_to="imageneslugar/")
texto = models.CharField(max_length=400)
The thing is that when i create a new instance of this model and
assign instance.archivo to a file then instance.archivo.path does
worked like a charm. thx.
On Sep 18, 5:21 pm, gnijholt <[EMAIL PROTECTED]> wrote:
> I had to define the callable function before it gets called.
> So in your case I'd put it right under class Product(models.model)
> Then you should be able to just do:
> upload_to=get_image_path
>
> Cheers,
>
>
I had to define the callable function before it gets called.
So in your case I'd put it right under class Product(models.model)
Then you should be able to just do:
upload_to=get_image_path
Cheers,
On Sep 18, 10:20 pm, lingrlongr <[EMAIL PROTECTED]> wrote:
> According to the documentation, I ca
According to the documentation, I can make the upload_to argument a
callable (which must receive 2 args):
class Product(models.Model):
number = models.CharField(max_length=20)
image = models.ImageField(upload_to='')
I want to build a path so it looks like this (with respect to
MEDIA_URL)
And in case anyone is interested in the results (a demo django gallery
application) have a look here:
http://saschashideout.de/wiki/DjangoGalleryTutorial/
--~--~-~--~~~---~--~~
You received this message because you are subscribed to the Google Groups
"Django use
Thanks, this is quite a good idea, the more I think about it the more
I like it, because Images have only one Gallery, but I can change the
gallery for every image if I want, which would lead to the problem of
moving the image file from one directory to another. Your proposal
seems to be the bette
Am Wed, 04 Jul 2007 01:19:16 -0700
schrieb lastmohican <[EMAIL PROTECTED]>:
> The Problem is, I want a dynamic image
> storage path, something like MEDIA_ROOT+"images/" wont work with
> hundreds of galleries, [...]
Since your primary concern seems to be too many files in one directory,
did you c
Hello everybody,
I want to create a gallery app and I have a slight problem, my
models.py
code looks like this:
...
class Gallery(models.Model):
slug = models.SlugField(...)
...
class Image(models.Model):
...
gallery = models.ForeignKey(...)
image = models.ImageField(
uplo
timster wrote:
> Should the super() method work even though I'm not using MR?
No. super().save() only works in MR. You'll need the _pre_save() and
_post_save() hooks in .91 or trunk.
--
--Max Battcher--
http://www.worldmaker.net/
"I'm gonna win, trust in me / I have come to save this world /
timster wrote:
> Should the super() method work even though I'm not using MR?
No. super().save() only works in MR. You'll need the _pre_save() and
_post_save() hooks in .91 or trunk.
--
--Max Battcher--
http://www.worldmaker.net/
"I'm gonna win, trust in me / I have come to save this world /
Thanks for the reply Ian.
I seem to have solved this for the most part. The second message in
that thread was very helpful. I was able to use that solution almost
exactly. I added the following save() method to my model:
def save(self):
if self.image:
import shutil
from os im
On 4/4/06, timster <[EMAIL PROTECTED]> wrote:
> What I would like to do is store the image in a subfolder corresponding
> to the gallery_id. If "Subaru" is gallery_id 1 and "Audi" is gallery_id
> 2, I would want it to look like this:
>
> gallery/1/subaru1.jpg
> gallery/2/audi1.jpg
>
> Is this poss
I'm trying to create a simple image gallery. Here's what my models look
like. It's very simple, a gallery can have one or more images.
class Gallery(meta.Model):
name = meta.CharField(maxlength=50)
class Image(meta.Model):
gallery = meta.ForeignKey(Gallery)]
image = meta.ImageField(u
21 matches
Mail list logo
