Makes sense, I probably should have realized this from the beginning.
Appreciate the help.
On Monday, September 20, 2021 at 6:17:05 PM UTC-5 Sean wrote:
> Add a property to count Y for each column
>
> Or add a computed field that tallies that on save
>
> For reference
>
>
> https://stackoverflo
Sum, Count works when we try to get sum/count for a column,
for this case you can try something like this
def get_yes_count_by(id):
user = User.objects.get(id=id)
count = 0
for val in user.__dict__.values():
if val == "YES":
count += 1
return count
On Tue, 21 S
Add a property to count Y for each column
Or add a computed field that tallies that on save
For reference
https://stackoverflow.com/questions/1454727/do-properties-work-on-django-model-fields
On Mon, Sep 20, 2021 at 6:41 PM J.T. wrote:
> I have a model called member that has 6 columns: ID,
Tosin,
Please refer below. This should help.
https://docs.djangoproject.com/en/2.2/topics/db/queries/
best,
matt
On Thu, Sep 12, 2019 at 10:01 AM Tosin Ayoola
wrote:
> Halo guyz I actually need a little help, I'm working on a airline tickets
> reservations project using postman Api that requir
On 03/11/2010, at 2:10 AM, Prashanth wrote:
>
>
> On Tue, Nov 2, 2010 at 9:35 AM, Itai Tavor wrote:
> Hi people,
>
> Given this:
>
> class Category(models.Model):
>name = models.CharField()
>parent = models.ForeignKey('self', blank=True, null=True)
>tree_path = models.CharField() #
On Tue, Nov 2, 2010 at 9:35 AM, Itai Tavor wrote:
> Hi people,
>
> Given this:
>
> class Category(models.Model):
>name = models.CharField()
>parent = models.ForeignKey('self', blank=True, null=True)
>tree_path = models.CharField() # //...///
>
> class Item(models.Model):
>name = m
Took me awhile to wrap my head around it, but that got me where I need
to be, thank you.
On Jun 15, 3:41 pm, Streamweaver wrote:
> A custom method in the model could work fine.
>
> If you set a 'get_latest_by' in the Entry Model
> ->http://www.djangoproject.com/documentation/models/get_latest/
A custom method in the model could work fine.
If you set a 'get_latest_by' in the Entry Model ->
http://www.djangoproject.com/documentation/models/get_latest/
Then in the Blog Model define something like:
def last_is_foo(self):
if Entry.objects.latest().title == 'foo'
return True
Close. That returns all entrys that have title foo. I need all blogs
*where the last entry* has title foo. I think I need to
leverage .extra(where=something) but not having much luck.
On Jun 15, 2:36 pm, Streamweaver wrote:
> If I understand your question right there is an example that cover
If I understand your question right there is an example that covers
this exact situation that may help at
http://docs.djangoproject.com/en/dev/topics/db/queries/#lookups-that-span-relationships
I believe it could come out to something like this:
e = Entry.objects.filter(title__exact='foo')
blo
Thats good, but now I need to turn those id's back into content types.
>From there I need to get the model's verbose plural name also.
The idea is a view where the types of assets are listed, Rooms,
Projectors, etc.
BTW, if you couldn't tell from the sql statement before, this is a
booking appli
On Jun 5, 6:56 pm, "eric.frederich" wrote:
> How can I get distinct content_types from a model like this?...
>
> In sqlite I did the following and it worked...
> sqlite> select distinct content_type_id from booking_managedasset;
>
> class ManagedAsset(models.Model):
> content_type = models.
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