Sorry about the confusing title, but I don't know how to describe it better.
I need to run a model-function on the object I am editing using PUT in Django-Rest-Framework, which uses some of the new data from the PUT to calculate some new values it should save in the same model.
Example:
An item with {'amount': 2, 'price': 0, 'total': 0} is already stored in the database.
I am then updating price to 1 using a normal PUT request using django-rest-framework.
The model have a helperfunction called update_total() which I need to call to update the total field in the database (to, in this case 2 (2*1)).
The item is updated in the database, but the response returned from django-rest-framework is still showing total=0. After getting the object on new, total will be 2 as expected.
I need the response to be 2 in the response from the PUT, not after a regrab of the object. But how?
I have tried several things (which all doesn’t work):
Updating attrs in a validator to the new value.
Using post_save() in ListCreateAPIView to update the data.
Using pre_save() in ListCreateAPIView
Updating instance in restore_object() (even though it isn't for this purpose)
Does this look like a bug? Or is there another trick?
I kinda found a solution, but it feels somewhat dirty..
In my serializers restore_object I put code like this:
new_values = instance.update_counters()
for k, v in new_values.items():
self.data[k] = v
and in my models update_counters() function, I am returning a dict of what I changed..
Related
I've been using the get_or_create method with MongoEngine in a Django app. Today, I noticed there were a few duplicate entries. I came across this in the MongoEngine API Reference for get_or_create:
This requires two separate operations and therefore a race condition exists. Because there are no transactions in mongoDB other approaches should be investigated, to ensure you don’t accidentally duplicate data when using this method. This is now scheduled to be removed before 1.0
Should I be using something like this?:
from models import Post
post = Post(name='hello')
try:
Posts.objects.get(name=post.name)
print "exists"
except:
post.save()
print "saved"
Will that solve my problem?
Is there a better way?
To perform an upsert use the following syntax:
Posts.objects(name="hello").update(set__X=Y, upsert=True)
That will add a post with the name "hello" and where X = Y if it doesn't already exist, otherwise it will update an existing post just setting X = Y.
If you need pass a dictionery, can do this:
post = Post.objects(name="hello").first() or Post(name="hello")
then you can update with something like this:
# data = dictionary_with_data
for field, value in data.items():
post[field] = value
post.save()
I understand that Django want to generate forms automatically so you don't have to do so in your template, and I do understand that many people find it cool.
But I have specific requirements and I have to write my forms on my own. I just need something to parse the data, be it a form submitted using a user interface, or an API request, or whatever.
I tried to use ModelForm, but it doesn't seem to work as I want it to work.
I'd like to have something with the following behavior:
possibility to specify the model of the object I am going to create/update
possibility to specify an object in case of an update
possibility to provide new data in a dictionary
if I am creating a new object, missing fields in my data should be replaced by their default values as specified in my model definition
if I am updating an existing object, missing fields in my data should be replaced by the current values of the object I am updating. Another way of saying is, do not update values that are missing in my data dictionary.
data validation should be performed before calling save(), and it should throw a ValidationError with the list of erroneous fields and errors.
Currently, I prefer to do everything manually :
o = myapp.models.MyModel() # or o = myapp.Models.MyModel.objects.get(pk = data['pk'])
o.field1 = data['field1']
o.field2 = data['field2']
…
o.full_clean()
o.save()
It would be nice to have a shortcut :
o = SuperCoolForm(myapp.models.MyModel, data)
o.save()
Do you know if Django does provide a solution for this or am I asking too much?
Thank you!
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.
Im playing a little bit with heavy-client app.
Imagine I have this model:
class Category(models.Model):
name = models.CharField(max_length=30)
color = models.CharField(max_length=9)
Im using knockoutjs (but I guess this is not important). I have a list (observableArray) with categories and I want to create a new category.
I create a new object and I push it to the list. So far so good.
What about saving it on my db? Because I'm using tastypie I can make a POST to '/api/v1/category/' and voilà, the new category is on the DB.
Ok, but... I haven't refresh the page, so... if I want to update the new category, how I do it?
I mean, when I retrieve the categories, I can save the ID so I can make a put to '/api/v1/category/id' and save the changes, but... when I create a new category, the DB assign a id to it, but my javascript doesn't know that id yet.
in other words, the workflow is something like:
make a get > push the existing objects (with their ids) on a list > create a new category > push it on the list > save the existing category (the category doesnt have the id on the javacript) > edit the category > How I save the changes?
So, my question is, what's the common path? I thought about sending the category and retrieving the id somehow and assign it to my object on js to be able to modify it later. The problem is that making a POST to the server doesn't return anything.
In the past I did something like that, send the object via post, save it, retrieve it and send it back, on the success method retrieve the id and assign it to the js object.
Thanks!
Tastypie comes with an always_return_data option for Resources.
When always_return_data=True for your Resource, the API always returns the full object event on POST/PUT, so that when you create a new object you can get the created ID on the same request.
You can then just read the response from your AJAX and decode the JSON (i dont know about knockout yet).
see the doc : http://readthedocs.org/docs/django-tastypie/en/latest/resources.html?highlight=always_return_data#always-return-data
Hope this helps
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)