When a user updates an invoice form, i want to create a new invoice record with the updated attributes, but also change one or two fields of the old record and save it, too.
How would the outline of a controller action look like which could accomplish this?
Instead of a controller action i put the code in the model, using callbacks:
before_save do |rec|
if !rec.new_record?
attrb = rec.attributes.delete_if{|k, v| ["id"].include? k }
Book.create(attrb)
rec.restore_attributes
rec.year = rec.year + 2 # some custom change
true
end
end
I keep all attributes unless the 'id' (otherwise i get an error) for create a new record with the new attributes.
Then i restore the attributes of the existing record. I do some custom change before saving.
I am rather new with Rails but this seems pretty straightforward. As you mention the user is 'updating" an invoice, your controller view has probably been passed all the data available to the user for further change.
When submitting the form, your update action can easily update the current record data, as well as creating a new one on top of this
Though as it is automated, you need to make clear:
if a new invoice record is created each time an invoice record is
updated (thi can create a lot of copies of the same invoice)
how you make the old record an archive to avoid duplicates
can the 'additional" amendments be automated and easily processed through an algorithm...
Nested attributes made things a bit tricky. So in order to create new instances I had to use the dup method for both the resource and its nested items.
Generally, it is advisable to keep the controllers slim and make the models fat. Nevertheless, I have decided to include this code into my Invoices controller:
def revise_save
#contact = Contact.find(params[:contact_id])
#invoice = #contact.invoices.find(params[:invoice_id])
#invoice_old = #invoice.dup
#invoice.invoice_items.each do |item|
#invoice_old.invoice_items << item.dup
end
#invoice.datum = DateTime.now.to_date
# archive old invoice
# #invoice_old. ...
#invoice_old.save
# make old new invoice
#invoice.datum = Time.now
# ...
#invoice.update(invoice_params)
redirect_to invoices_path
end
Note that in this solution the currently edited (original) invoice becomes the new invoice, the old one is paradoxically created anew.
Thanks to #iwan-b for pointing me in the right direction.
Related
I think what I'm trying to achieve is not hard, but I have no clue how to do it hehehehe !
Basically what I need is the feature that we have in Django Admin, when you are creating a new object, if you have a Foreign Key, you can add new data (opening a pop-up), save it and then the select box updates automatically.
What I have is this form:
I know that would be easy to do it with some Javascript, but my point is, Django has some rules, and as far I know, I can't add new data to a form already created, right? Otherwise Django won't validate this form. How could I achieve this?
PS: "Local" is the select box where I want to add new data. The user should be able to create a new Local on this page, instead of going to another page to do it. Thanks :)
Here your question:
I can't add new data to a form already created, right? Otherwise Django won't validate this form. How could I achieve this?
Then the answer:
you are right, django will check values match form value rules. But:
realize that your main form is invoked for twice: on GET and on POST. Between both form executions you make changes on database values trhough your new form. That means that in second main form invocation the value added to database is available:
field1 = forms.ModelChoiceField(queryset= ***1*** )
***1***: on second invocation new value is already available on field1.
Then, you don't should to be afraid about this subject, the new value will be available on form on your main form POST request.
Nothing wrong with updating the value using javascript as long the key in your new combo box has the right key in the database then it should be ok.
Call this function after you saved the last entry.
function refreshLocal(){
$.get(window.location.href, '', function(html){
// change the id to the local combox's id
var serverLocalDropBox = $(html).find('#id_local');
if (serverLocalDropBox.length){
$('#id_local').replaceWith(serverLocalDropBox);
}
})
}
If you don't want to use javascript solution, you can post the form with refresh flag and on the server side if you see that flag just don't validate and return the form as is. Since you have a new entry in the foreignkey it will automatically update the queryset to include the new entry.
function serverRefreshLocal(){
var $form = $('#your_form_id');
$form.append('<input type="hidden" name="refresh" value="true" />');
// you can use ajax submit and ajax refresh here if you don't want to leave the page
$form.submit();
}
// Server Side
def your_form_post_view(request):
if request.POST.get('refresh', 'false') == 'true':
# initial is the trick to save user input
your_form = YourForm(initial=request.POST)
context = {
'form': your_form,
}
return render(request, 'your_template.html', context)
# your view code goes here
I've just added a new column to an existing table in my database:
class AddMoveableDateToDocument < ActiveRecord::Migration
def change
add_column :documents, :moveable_date, :datetime
end
end
In my Rails model I want the moveable_date attribute to be set to a default value upon creation, and the application will be able to change this date later. So, something like:
class Document < ActiveRecord::Base
before_create :set_moveable_date
def set_moveable_date
self.moveable_date ||= self.created_at
end
end
Now, the existing models that are already saved into the database will not have this moveable_date set yet (the value is nil). How do I run through all my existing models and populate the moveable_date attribute with its default value? What is the easiest/best practice way? Can be in the application code itself, in the console, in the terminal, or otherwise. Thanks!
You will get a lot of opinionated answers on this one. Some will suggest the console, some will suggest a one-time rake task.
I would suggest doing it as part of the migration that adds the column. After adding the column, you can run Document.reset_column_information so that the Rails app picks up on your new column, and then iterate through the existing document records and set the moveable date as appropriate.
Or if it's as simple as setting the moveable date to the created_at date, you can use something like Document.update_all("moveable_date = created_at") instead of iterating over them.
That's a good suggestion!
Another way is to add the line before_save :set_moveable_date to your model. It won't accomplish the transition immediately, but if your data is updated on a regular basis, it'd work.
I have some nested Ember Data models week->hasMany('workout')->hasMany('exercises')
In my app I need to create an entire new week. I accomplish this by
var newWeek = WT.PlanWeek.store.createRecord('planWeek', attributes );
// create a couple new workouts
newWorkout = WT.PlanWorkout.store.createRecord('planWorkout', attributes);
newWeek.get('workouts').pushObject(newWorkout);
// repeat...
// create exercises on the workouts the same way and push them on the the workout
newExercise = WT.PlanWeek.store.createRecord('planExercise', attributes);
newWorkout.get('exercises').pushObject(newExercise);
newWorkout.save();
I have a couple problems with this save.
I save the both the workout and the child exercises in the workout API call - when the call returns this appends a second set of exercises to the workout instead of replacing the existing exercises (even though I set workout.exercises to the newly persisted exercises in the serializer extractSingle).
When the server returns it doesn't update the workouts in the week with the new ids. I end up with the new persisted workouts that correctly belongTo the week, but week.get('workouts') still points to the unpersisted workouts with no id.
Any ideas or suggestions?
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)