I'm using sqlalchemy and have two models, Article and Tag, it's a many-to-many relation.
When I add articles using Flask-Admin, I want just part of tags available (related on user permission) instead of all tags.
any idea? Thanks
Probably the best way to do this is to use dynamic relationship loaders. Simply use lazy='dynamic' in your relationship definition:
posts = relationship(Post, lazy="dynamic")
This returns you a query object instead of a collection of objects, so you can then query it directly:
posts = jack.posts.filter(Post.headline=='this is a post')
You could also achieve what you want with discriminator columns or something, but that is likely overkill.
sounds like you need ModelView.get_query:
class MyView(ModelView):
def get_query(self,*args,**kwargs):
return super(MyView,self).get_query(*args,**kwargs).filter_by(current_user.can_view=True)
Related
I have some models using django-polymorphic-model
eg.
class Article(PolymorphicModel):
...
class Blog(Article):
tags = ...
class Story(Article):
publish = ...
Normally if I get all articles, I just do Article.objects.all(), however what if I want to get all articles that tags are empty? If I do Articles.objects.filter(tags__isnull=True) it will break because other models don't have this field, I would like to include Story entries too, do I really have to split into 2 different queries and combine again?
OK After some digging for the documentation through issues, here is how to do it
Articles.objects.filter(Blog___tags__isnull=True)
Hope you found an answer before.
Articles.objects.filter(Q(Blog___tags__isnull=True)|Q(Story__tags__isnull=True))
I was wondering what is the correct approach,
Do I create HiddenInput fields in my ModelForm and from the
View I pass in the primaryKey for the models I am about to edit into
the hiddenInput fields and then grab those hiddenInput fields from
the AJAX script to use it like this?
item.load(
"/bookmark/save/" + hidden_input_field_1,
null,
function () {
$("#save-form").submit(bookmark_save);
}
);
Or is there is some more clever way of doing it and I have no idea?
Thanks
It depends upon how you want to implement.
The basic idea is to edit 1. you need to get the existing instance, 2. Save provided information into this object.
For #1 you can do it multiple ways, like passing ID or any other primary key like attribute in url like http://myserver/edit_object/1 , Or pass ID as hidden input then you have to do it through templates.
For #2, I think you would already know this. Do something like
inst = MyModel.objects.get(id=input_id) # input_id taken as per #1
myform = MyForm(request.POST, instance=inst)
if myform.is_valid():
saved_inst = myform.save()
I just asked in the django IRC room and it says:
since js isn't processed by the django template engine, this is not
possible.
Hence the id or the object passed in from django view can't be accessed within AJAX script.
I've been using the built-in Django comments system which has been working great. On a particular page I need to list the latest X comments which I've just been fetching with:
latest_comments =
Comment.objects.filter(is_public=True, is_removed=False)
.order_by('submit_date').reverse()[:5]
However I've now introduced a Boolean field 'published' into the parent object of the comments, and I want to include that in the query above. I've tried using the content_type and object_pk fields but I'm not really getting anywhere. Normally you'd do something like:
Comment.objects.filter(blogPost__published=True)
But as it is not stored like that I am not sure how to proceed.
posts_ids = BlogPost.objects.filter(is_published=True).values_list('id', flat=True) #return [3,4,5,...]
ctype = ContentType.objects.get_for_model(BlogPost)
latest_comments = Comment.objects.filter(is_public=True, is_removed=False, content_type=ctype, content_object__in=posts_ids).order_by('-submit_date')[:5]
Comments use GenericForeignKey to store the relation to parent object. Because of the way generic relations work related lookups using __<field> syntax are not supported.
You can accomplish the desired behaviour using the 'in' lookup, however it'll require lot of comparisons when there'll be a lot of BlogPosts.
ids = BlogPost.objects.filter(published=True).values_list('id', flat=True) # Get list of ids, you would probably want to limit number of items returned here
content_type = ContentType.objects.get_for_model(BlogPost) # Becasue we filter only comments for BlogPost
latest_comments = Comment.objects.filter(content_type=content_type, object_pk__in=ids, is_public=True, is_removed=False, ).order_by('submit_date').reverse()[:5]
See the Comment model doc for the description of all fields.
You just cannot do that in one query. Comments use GenericForeignKey. Documentation says:
Due to the way GenericForeignKey is implemented, you cannot use such
fields directly with filters (filter() and exclude(), for example) via
the database API.
I have a lot of objects to save in database, and so I want to create Model instances with that.
With django, I can create all the models instances, with MyModel(data), and then I want to save them all.
Currently, I have something like that:
for item in items:
object = MyModel(name=item.name)
object.save()
I'm wondering if I can save a list of objects directly, eg:
objects = []
for item in items:
objects.append(MyModel(name=item.name))
objects.save_all()
How to save all the objects in one transaction?
as of the django development, there exists bulk_create as an object manager method which takes as input an array of objects created using the class constructor. check out django docs
Use bulk_create() method. It's standard in Django now.
Example:
Entry.objects.bulk_create([
Entry(headline="Django 1.0 Released"),
Entry(headline="Django 1.1 Announced"),
Entry(headline="Breaking: Django is awesome")
])
worked for me to use manual transaction handling for the loop(postgres 9.1):
from django.db import transaction
with transaction.atomic():
for item in items:
MyModel.objects.create(name=item.name)
in fact it's not the same, as 'native' database bulk insert, but it allows you to avoid/descrease transport/orms operations/sql query analyse costs
name = request.data.get('name')
period = request.data.get('period')
email = request.data.get('email')
prefix = request.data.get('prefix')
bulk_number = int(request.data.get('bulk_number'))
bulk_list = list()
for _ in range(bulk_number):
code = code_prefix + uuid.uuid4().hex.upper()
bulk_list.append(
DjangoModel(name=name, code=code, period=period, user=email))
bulk_msj = DjangoModel.objects.bulk_create(bulk_list)
Here is how to bulk-create entities from column-separated file, leaving aside all unquoting and un-escaping routines:
SomeModel(Model):
#classmethod
def from_file(model, file_obj, headers, delimiter):
model.objects.bulk_create([
model(**dict(zip(headers, line.split(delimiter))))
for line in file_obj],
batch_size=None)
Using create will cause one query per new item. If you want to reduce the number of INSERT queries, you'll need to use something else.
I've had some success using the Bulk Insert snippet, even though the snippet is quite old.
Perhaps there are some changes required to get it working again.
http://djangosnippets.org/snippets/446/
Check out this blog post on the bulkops module.
On my django 1.3 app, I have experienced significant speedup.
bulk_create() method is one of the ways to insert multiple records in the database table. How the bulk_create()
**
Event.objects.bulk_create([
Event(event_name="Event WF -001",event_type = "sensor_value"),
Entry(event_name="Event WT -002", event_type = "geozone"),
Entry(event_name="Event WD -001", event_type = "outage") ])
**
for a single line implementation, you can use a lambda expression in a map
map(lambda x:MyModel.objects.get_or_create(name=x), items)
Here, lambda matches each item in items list to x and create a Database record if necessary.
Lambda Documentation
The easiest way is to use the create Manager method, which creates and saves the object in a single step.
for item in items:
MyModel.objects.create(name=item.name)
I'd like to have access to one my model field verbose_name.
I can get it by the field indice like this
model._meta._fields()[2].verbose_name
but I need to get it dynamically. Ideally it would be something like this
model._meta._fields()['location_x'].verbose_name
I've looked at a few things but I just can't find it.
For Django < 1.10:
model._meta.get_field_by_name('location_x')[0].verbose_name
model._meta.get_field('location_x').verbose_name
For Django 1.11 and 2.0:
MyModel._meta.get_field('my_field_name').verbose_name
More info in the Django doc
The selected answer gives a proxy object which might look as below.
<django.utils.functional.__proxy__ object at 0x{SomeMemoryLocation}>
If anyone is seeing the same, you can find the string for the verbose name in the title() member function of the proxy object.
model._meta.get_field_by_name(header)[0].verbose_name.title()
A better way to write this would be:
model._meta.get_field(header).verbose_name.title()
where header will be the name of the field you are interested in. i.e., 'location-x' in OPs context.
NOTE: Developers of Django also feel that using get_field is better and thus have depreciated get_field_by_name in Django 1.10. Thus I would suggest using get_field no matter what version of Django you use.
model._meta.get_field_by_name('location_x')[0].verbose_name
You can also use:
Model.location_x.field.verbose_name
Model being the class name. I tested this on my Animal model:
Animal.sale_price.field.verbose_name
Animal.sale_price returns a DeferredAttribute, which has several meta data, like the verbose_name
Note: I'm using Django 3.1.5
If you want to iterate on all the fields you need to get the field:
for f in BotUser._meta.get_fields():
if hasattr(f, 'verbose_name'):
print(f.verbose_name)
# select fields for bulk_update : exclude primary key and relational
fieldsfields_to_update = []
for field_to_update in Model._meta.get_fields():
if not field_to_update.many_to_many and not field_to_update.many_to_one and not field_to_update.one_to_many and not field_to_update.one_to_one and not field_to_update.primary_key and not field_to_update.is_relation :
fields_to_update = fields_to_update + [field_to_update.name]
Model.objects.bulk_update(models_to_update , fields_to_update)