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Ember nested route. Load more models

- by user3568719

JsBin http://jsbin.com/EveQOke/153/ I know how to load more then one model to a route, using Ember.RSVP.hash. (see Jsbin Children menu). I use dynamic part to access one elem from a collection children/1. But i cant load more models to a nested resource. In my example i want to populate all the toys for a select, not just list the toys of the child. I have tried to access the model of the route children App.ChildRoute = Ember.Route.extend({ model: function(param){ return Ember.RSVP.hash({ allToys: this.modelFor("children"), child:this.store.find('child', param.child_id) }); } }); and use its model's toy property (since there have already loaded all of the toys) child.hbs <h4>All avaiable toys</h4> <table> {{#each toy in model.allToys.toys}} <tr> <td>{{toy.name}}</td> </tr> {{/each}} </table>

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How to fetch populated associated models in CakePHP when calling read()

- by Code Commander

I have the following Models: class Site extends AppModel { public $name = "Site"; public $useTable = "site"; public $primaryKey = "id"; public $displayField = 'name'; public $hasMany = array('Item' => array('foreignKey' => 'siteId')); public function canView($userId, $isAdmin = false) { if($isAdmin) { return true; } return array_key_exists($this->id, $allowedSites); } } and class Item extends AppModel { public $name = "Item"; public $useTable = "item"; public $primaryKey = "id"; public $displayField = 'name'; public $belongsTo = array('Site' => array('foreignKey' => 'siteId')); public function canView($userId, $isAdmin = false) { // My problem appears to be the next line: return $this->Site->canView($userId, $isAdmin); } } In my controller I am doing something like this: $result = $this->Item->read(null, $this->request->id); // Verify permissions if(!$this->Item->canView($this->Session->read('userId'), $this->Session->read('isAdmin'))) { $this->httpCodes(403); die('Permission denied.'); } I notice that in Item->canView() $this->data['Site'] is populated with the column data from the site table. But it merely an array and not an object. On the other hand $this->Site is a Site object, but it has not been populated with the column data from the site table like $this->data. What is the proper way to have CakePHP get the associated model as the object and containing the data? Or am I going about this all wrong? Thanks!

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List of models in Model in MVC

- by arri

I have two models: class ModelIn{ public string FirstName { get; set; } public string LastName { get; set; } public string Address { get; set; } } class ModelOut{ public ModelOut(){ People = new List<ModelIn>();} public List<ModelIn> People { get; private set;} public string Country { get; set; } } And I have Controller editing ModelOut: public ActionResult People() { ... return View(SomeModelOutInstanceWith3People); } [HttpPost] public ActionResult(ModelOut m) { ... } In view I have sth like: <% using (Html.BeginForm()) { %> <%: Html.EditorFor(m => Model.Country) %> <% for(int i = 0; i < Model.People.Count; ++i){ %> <%: Html.EditorFor(m => Model.People[i].FirstName) %> <%: Html.EditorFor(m => Model.People[i].LastName) %> <%: Html.EditorFor(m => Model.People[i].Address) %> <% } %> <input type="submit" /> <% } %> It works all OK, but in post action I have empty ModelOut m. I can see in logs that data is sent correctly. I have tried everything, nothing works.

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Initialize child models at model creation

- by Antoine

I have a model Entree which belongs to a model Vin, which itself belongs to a model Producteur. On the form for Entree creation/edition, I want to allow the user to define the attributes for parent Vin and Producteur to create them, or retrieve them if they exist (retrieval based on user input). For now I do the following in Entree new and edit actions: @entree = Entree.new @entree.vin = Vin.new @entree.vin.producteur = Producteur.new and use fields_for helper in the form,and that works. But I intend to have much more dependencies with more models, so I want to keep it DRY. I defined a after_initialize callback in Vin model which does the producteur initialization: class Vin < ActiveRecord::Base after_initialize :vin_setup def vin_setup producteur = Producteur.new end end and remove the producteur.new from the controller. However, get an error on new action: undefined method `model_name' for NilClass:Class for the line in the form that says <%= fields_for @entree.vin.producteur do |producteur| %> I guess that means the after_initialize callback doesn't act as I expect it. Is there something I'm missing? Also, I get the same error if I define a after_initialize method in the Vin model instead of definiing a callback.

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ActiveRecord Validations for Models with has_many, belongs_to associations and STI

- by keruilin

I have four models: User Award Badge GameWeek The associations are as follows: User has many awards. Award belongs to user. Badge has many awards. Award belongs to badge. User has many game_weeks. GameWeek belongs to user. GameWeek has many awards. Award belongs to game_week. Thus, user_id, badge_id and game_week_id are foreign keys in awards table. Badge implements an STI model. Let's just say it has the following subclasses: BadgeA and BadgeB. Some rules to note: The game_week_id fk can be nil for BadgeA, but can't be nil for BadgeB. Here are my questions: For BadgeA, how do I write a validation that it can only be awarded one time? That is, the user can't have more than one -- ever. For BadgeB, how do I write a validation that it can only be awarded one time per game week?

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Using nHibernate to map two different data models to one entity model

- by Dan

I have two different data models that map to the same Car entity. I needed to create a second entity called ParkedCar, which is identical to Car (and therefore inherits from it) in order to stop nhibernate complaining that two mappings exists for the same entity. public class Car { protected Car() { IsParked = false; } public virtual int Id { get; set; } public bool IsParked { get; internal set; } } public class ParkedCar : Car { public ParkedCar() { IsParked = true; } //no additional properties to car, merely exists to support mapping and signify the car is parked } The only issue is that when I come to retrieve a Car from the database using the Criteria API like so: SessionProvider.OpenSession.Session.CreateCriteria<Car>() .Add(Restrictions.Eq("Id", 123)) .List<Car>(); The query brings back Car Entities that are from the ParkedCar data model. Its as if nhibernate defaults to the specialised entity. And the mappings are defiantly looking in the right place: <class name="Car" xmlns="urn:nhibernate-mapping-2.2" table="tblCar"> <class name="ParkedCar" xmlns="urn:nhibernate-mapping-2.2" table="tblParkedCar" > How do I stop this?

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How to validate properties across Models without repeating the validation logic

- by Mano

Hello All, I am building a ASP.NET Mvc app. I have a Data model say User public class user { public int userId {get; private set}; public string FirstName {get; set;} } The validation to be done is that the firstname cannot exceed 50 characters. I have another presentation model in which i have the property FirstName too. I do not want to repeat the validation logic in both the models. I want to have it in one place and that should be it. I can do it in a simpler way by adding a function which can be called while setting the property like private string firstName; public string FirstName { get { return firstName; } set { if (PropertyValidator.ValidName(value)) // assuming ValidName exists and it will throw an exception if the value is not valid { firstName = value; } } } But I am looking for something much simpler so that I do not need to add this for every property I need to have it validated. I looked at ValidationAttribute but then again I can validate this only from a controller (ModelState.IsValid). Since this model could be used by some other type of apps like console app, I could not choose that. But if there is a way to use the Mvc's ModelState.IsValid from outside of a controller, that would be awesome. Any suggestions are greatly appreciated. Thanks!!

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Passing models between controllers in MVC4

- by wtfsven

I'm in the middle of my first foray into web development with MVC4, and I've run into a problem that I just know is one of the core issues in web development in general, but can't seem to wrap my mind around it. So I have two models: Employee and Company. Employee has a Company, therefore /Employee/Create contains a field that is either a dropdown list for selecting an existing Company, or an ActionLink to /Company/Create if none exists. But here's the issue: If a user has to create a Company, I need to preserve the existing Employee model across both Views--over to /Company/Create, then back to /Employee/Create. So my first approach was to pass the Employee model to /Company/Create with a @Html.ActionLink("Add...", "Create", "Organization", Model , null), but then when I get to Company's Create(Employee emp), all of emp's values are null. The second problem is that if I try to come back to /Employee/Create, I'd be passing back an Employee, which would resolve to the POST Action. Basically, the whole idea is to save the state of the /Employee/Create page, jump over to /Company/Create to create a Company record, then back to /Employee/Create to finish creating the Employee. I know there is a well understood solution to this because , but I can't seem to phase it well enough to get a good Google result out of it. I've though about using Session, but it seems like doing so would be a bit of a kludge. I'd really like an elegant solution to this.

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How to refactor models without breaking WPF views?

- by Tim Murphy

I've just started learning WPF and like the power of databinding it presents; that is ignoring the complexity and confusion for a noob. My concern is how do you safely refactor your models/viewmodels without breaking the views that use them? Take the following snippet of a view for example: <Grid> <ListView ItemsSource="{Binding Contacts}"> <ListView.View> <GridView> <GridViewColumn Header="First Name" DisplayMemberBinding="{Binding Path=FirstName}"/> <GridViewColumn Header="Last Name" DisplayMemberBinding="{Binding Path=FirstName}"/> <GridViewColumn Header="DOB" DisplayMemberBinding="{Binding Path=DateOfBirth}"/> <GridViewColumn Header="# Pets" DisplayMemberBinding="{Binding Path=NumberOfPets}"/> <GridViewColumn Header="Male" DisplayMemberBinding="{Binding Path=IsMale}"/> </GridView> </ListView.View> </ListView> </Grid> The list is bound to the Contacts property, IList(Of Contact), of the windows DataSource and each of the properties for a Contact is bound to a GridViewColumn. Now if I change the name of the NumberOfPets property in the Contact model to PetCount the view will break. How do I prevent the view breaking?

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django modelformset - one form per related table row

- by Toby

Hello, I have two models: class Model1(): name = CharField() url = CharField() class Model2(): model1 = ForeignKey(Model1) user = ForeignKey(User) zzz = CharField() There are 5 rows for model1 in the database, these are fixed and will rarely change. I need to display a formset for model2 that allows users to enter the zzz value, the formset must always show one form per row in the model1 table, the label for each form in the formset must be the name of the related model1. If the user deletes a model2 in the formset the next time the page loads it will render an empty zzz value for that form and the user must be able to edit the previous zzz value - meaning it must be pre populated with all model2 rows associated with the user. The idea is to print each row in the model1 table as a form instead of the user selecting the related model1 name in a select box. I know its not that complicated, but I'm seriously stumped and keep going round in circles!! Many thanks in advance. Similar to http://stackoverflow.com/questions/298779/form-or-formset-to-handle-multiple-table-rows-in-django

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CodeIgniter & Datamapper as frontend, Django Admin as backend, database tables inconsistent

- by Rasiel

I created a database for a site i'm doing using Django as the admin backend. However because the server where the site is hosted on, won't be able to support Python, I find myself needing to do the front end in PHP and as such i've decided to use CodeIgniter along with Datamapper to map the models/relationship. However DataMapper requires the tables to be in a specific format for it to work, and Django maps its tables differently, using the App name as the prefix in the table. I've tried using the prefix & join_prefix vars in datamapper but still doesn't map them correctly. Has anyone used a combination of this? and if so how have the fixed the issue of db table names being inconsistent? Is there anything out there that i can use to make them work together?

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django forms doubt

- by webvulture

Here, I am a bit confused with forms in Django. I have information for the form(a poll i.e the poll question and options) coming from some db_table - table1 or say class1 in models. Now the vote from this poll is to be captured which is another model say class2. So, I am just getting confused with the whole flow of forms, here i think. How will the data be captured into the class2 table? I was trying something like this. def blah1() get_data_from_db_table_1() x = blah2Form() render_to_response(blah.html,{...})

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Using multiple column layout with HTML 5 and CSS 3

- by nikolaosk

This is going to be the fourth post in a series of posts regarding HTML 5. You can find the other posts here , here and here.In this post I will provide a hands-on example with HTML 5 and CSS 3 on how to create a page with multiple columns and proper layout.I will show you how to use CSS 3 to create columns much easier than relying on DIV elements and the float CSS rule.I will also show you how to use browser-specific prefix rules (-ms for Internet Explorer and -moz for Firefox ) for browsers that do not fully support CSS 3.In order to be absolutely clear this is not (and could not be) a detailed tutorial on HTML 5. There are other great resources for that.Navigate to the excellent interactive tutorials of W3School.Another excellent resource is HTML 5 Doctor.Two very nice sites that show you what features and specifications are implemented by various browsers and their versions are http://caniuse.com/ and http://html5test.com/. At this times Chrome seems to support most of HTML 5 specifications.Another excellent way to find out if the browser supports HTML 5 and CSS 3 features is to use the Javascript lightweight library Modernizr.In this hands-on example I will be using Expression Web 4.0.This application is not a free application. You can use any HTML editor you like.You can use Visual Studio 2012 Express edition. You can download it here.I will create a simple page with information about HTML 5, CSS 3 and JQuery. This is the full HTML 5 code. <!DOCTYPE html><html lang="en"> <head> <title>HTML 5, CSS3 and JQuery</title> <meta http-equiv="Content-Type" content="text/html;charset=utf-8" > <link rel="stylesheet" type="text/css" href="style.css"> </head> <body> <div id="header"> <h1>Learn cutting edge technologies</h1> <p>HTML 5, JQuery, CSS3</p> </div> <div id="main"> <div id="mainnews"> <div> <h2>HTML 5</h2> </div> <div> <p> HTML5 is the latest version of HTML and XHTML. The HTML standard defines a single language that can be written in HTML and XML. It attempts to solve issues found in previous iterations of HTML and addresses the needs of Web Applications, an area previously not adequately covered by HTML. </p> <div class="quote"> <h4>Do More with Less</h4> <p> jQuery is a fast and concise JavaScript Library that simplifies HTML document traversing, event handling, animating, and Ajax interactions for rapid web development. </p> </div> <p> The HTML5 test(html5test.com) score is an indication of how well your browser supports the upcoming HTML5 standard and related specifications. Even though the specification isn't finalized yet, all major browser manufacturers are making sure their browser is ready for the future. Find out which parts of HTML5 are already supported by your browser today and compare the results with other browsers. The HTML5 test does not try to test all of the new features offered by HTML5, nor does it try to test the functionality of each feature it does detect. Despite these shortcomings we hope that by quantifying the level of support users and web developers will get an idea of how hard the browser manufacturers work on improving their browsers and the web as a development platform.</p> </div> </div> <div id="CSS"> <div> <h2>CSS 3 Intro</h2> </div> <div> <p> Cascading Style Sheets (CSS) is a style sheet language used for describing the presentation semantics (the look and formatting) of a document written in a markup language. Its most common application is to style web pages written in HTML and XHTML, but the language can also be applied to any kind of XML document, including plain XML, SVG and XUL. </p> </div> </div> <div id="CSSmore"> <div> <h2>CSS 3 Purpose</h2> </div> <div> <p> CSS is designed primarily to enable the separation of document content (written in HTML or a similar markup language) from document presentation, including elements such as the layout, colors, and fonts.[1] This separation can improve content accessibility, provide more flexibility and control in the specification of presentation characteristics, enable multiple pages to share formatting, and reduce complexity and repetition in the structural content (such as by allowing for tableless web design). </p> </div> </div> </div> <div id="footer"> <p>Feel free to google more about the subject</p> </div> </body> </html> The markup is very easy to follow. I have used some HTML 5 tags and the relevant HTML 5 doctype.The CSS code (style.css) follows body{ line-height: 30px; width: 1024px; background-color:#eee; } p{ font-size:17px; font-family:"Comic Sans MS" } p,h2,h3,h4{ margin: 0 0 20px 0; } #main, #header, #footer{ width: 100%; margin: 0px auto; display:block; } #header{ text-align: center; border-bottom: 1px solid #000; margin-bottom: 30px; } #footer{ text-align: center; border-top: 1px solid #000; margin-bottom: 30px; } .quote{ width: 200px; margin-left: 10px; padding: 5px; float: right; border: 2px solid #000; background-color:#F9ACAE; } .quote :last-child{ margin-bottom: 0; } #main{ column-count:2; column-gap:20px; column-rule: 1px solid #000; -moz-column-count: 2; -webkit-column-count: 2; -moz-column-gap: 20px; -webkit-column-gap: 20px; -moz-column-rule: 1px solid #000; -webkit-column-rule: 1px solid #000; } All the rules in the css code are pretty simple. The layout is achieved with that CSS rule #main{ column-count:2; column-gap:20px; column-rule: 1px solid #000; -moz-column-count: 2; -webkit-column-count: 2; -moz-column-gap: 20px; -webkit-column-gap: 20px; -moz-column-rule: 1px solid #000; -webkit-column-rule: 1px solid #000; Do note the column-count,column-gap and column-rule properties. These properties make the two column layout possible.Please have a look at the picture below to see why I used prefixes for Chrome (webkit) and Firefox(moz).It clearly indicates that the CSS 3 column layout are not supported from Firefox and Chrome. Finally I test my simple HTML 5 page using the latest versions of Firefox,Internet Explorer and Chrome. In my machine I have installed Firefox 15.0.1.Have a look at the picture below to see how the page looks I have installed Google Chrome 21.0 in my machine.Have a look at the picture below to see how the page looks Have a look at the picture below to see how my page looks in IE 10. My page looks the same in all browsers. Hope it helps!!!

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Display/Edit fields across tables in a Django Form

- by jamida

I have 2 tables/models, which for all practical purposes are the same as the standard Author/Book example, but with the additional restriction that in my application each Author can only write 0 or 1 books. (eg, the ForeignKey field in the Book model is or can be a OneToOneField, and there may be authors who have no corresponding book in the Book table.) I would like to be able to display a form showing multiple books and also show some fields from the corresponding Author table (eg author_name, author_address). I followed the example of the inlineformset and but that doesn't even display the author name in the generated form.

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Decrease DB requests number from Django templates

- by Andrew

I publish discount offers for my city. Offer models are passed to template ( ~15 offers per page). Every offer has lot of items(every item has FK to it's offer), thus i have to make huge number of DB request from template. {% for item in offer.1 %} {{item.descr}} {{item.start_date}} {{item.price|floatformat}} {%if not item.tax_included %}{%trans "Without taxes"%}{%endif%} <a href="{{item.offer.wwwlink}}" >{%trans "Buy now!"%}</a> </div> <div class="clear"></div> {% endfor %} So there are ~200-400 DB requests per page, that's abnormal i expect. In django code it is possible to use select_related to prepopulate needed values, how can i decrease number of requests in template?

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Django ORM QuerySet intersection by a field

- by Sri Raghavan

These are the (pseudo) models I've got. Blog: name etc... Article: name blog creator etc User (as per django.contrib.auth) So my problem is this: I've got two users. I want to get all of the articles that the two users published on the same blog (no matter which blog). I can't simply filter the Article model by both users, because that would yield the set of Articles created by both users. Obviously not what I want. but can I filter somehow to get all of the articles where a field of the object matches between the two querysets?

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Can django's auth_user.username be varchar(75)? How could that be done?

- by perrierism

Is there anything wrong with running alter table on auth_user to make username be varchar(75) so it can fit an email? What does that break if anything? If you were to change auth_user.username to be varchar(75) where would you need to modify django? Is it simply a matter of changing 30 to 75 in the source code?: username = models.CharField(_('username'), max_length=30, unique=True, help_text=_("Required. 30 characters or fewer. Letters, numbers and @/./+/-/_ characters")) Or is there other validation on this field that would have to be changed or any other repercussions to doing so? See comment discussion with bartek below regarding the reason for doing it.

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Problems with contenttypes when loading a fixture in Django

- by gerdemb

I am having trouble loading Django fixtures into my MySQL database because of contenttypes conflicts. First I tried dumping the data from only my app like this: ./manage.py dumpdata escola > fixture.json but I kept getting missing foreign key problems, because my app "escola" uses tables from other applications. I kept adding additional apps until I got to this: ./manage.py dumpdata contenttypes auth escola > fixture.json Now the problem is the following constraint violation when I try to load the data as a test fixture: IntegrityError: (1062, "Duplicate entry 'escola-t23aluno' for key 2") It seems the problem is that Django is trying to dynamically recreate contenttypes with different primary key values that conflict with the primary key values from the fixture. This appears to be the same as bug documented here: http://code.djangoproject.com/ticket/7052 The problem is that the recommended workaround is to dump the contenttypes app which I'm already doing!? What gives? If it makes any difference I do have some custom model permissions as documented here: http://docs.djangoproject.com/en/dev/ref/models/options/#permissions

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How do I memoize expensive calculations on Django model objects?

- by David Eyk

I have several TextField columns on my UserProfile object which contain JSON objects. I've also defined a setter/getter property for each column which encapsulates the logic for serializing and deserializing the JSON into python datastructures. The nature of this data ensures that it will be accessed many times by view and template logic within a single Request. To save on deserialization costs, I would like to memoize the python datastructures on read, invalidating on direct write to the property or save signal from the model object. Where/How do I store the memo? I'm nervous about using instance variables, as I don't understand the magic behind how any particular UserProfile is instantiated by a query. Is __init__ safe to use, or do I need to check the existence of the memo attribute via hasattr() at each read? Here's an example of my current implementation: class UserProfile(Model): text_json = models.TextField(default=text_defaults) @property def text(self): if not hasattr(self, "text_memo"): self.text_memo = None self.text_memo = self.text_memo or simplejson.loads(self.text_json) return self.text_memo @text.setter def text(self, value=None): self.text_memo = None self.text_json = simplejson.dumps(value)

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Django: How to create a model dynamically just for testing

- by muhuk

I have a Django app that requires a settings attribute in the form of: RELATED_MODELS = ('appname1.modelname1.attribute1', 'appname1.modelname2.attribute2', 'appname2.modelname3.attribute3', ...) Then hooks their post_save signal to update some other fixed model depending on the attributeN defined. I would like to test this behaviour and tests should work even if this app is the only one in the project (except for its own dependencies, no other wrapper app need to be installed). How can I create and attach/register/activate mock models just for the test database? (or is it possible at all?) Solutions that allow me to use test fixtures would be great.

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Complex Django filter question

- by HWM-Rocker

Lets say I have this class (simplified): class Tag (...): children = models.ManyToManyField(null=True, symmetrical=False) Now I already implemented the functions get_parents, get_all_ancestors. Is there a nice pythonic way to just the top level tags? If I had designed my Tags differently (to point to the parents instead) I would just make get_all_parents().filter(children=null). My first thought is to create a new function that will go recursively through all parents and save those that has none. But is there a possibility with filters or Query-objects to do the same (with fewer lines of code)? Thanks for your help. [edit] When it is finished, it should be a hierarchical tagging system. Each tag can have children, parents, but only the children are saved. I want to get all the top level tags, that point through many children / childrens children to my tag.

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Problem with migrating a model in ruby

- by Shreyas Satish

I run script/generate model query edit query.rb in models.. class Query < ActiveRecord::Base #I even tried Migrations instead of Base def sef.up create table :queries do|t| t.string :name end end def self.down drop_table :queries end end ,run rake db:migrate. and what I see in db is this: mysql> desc queries; +------------+----------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +------------+----------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | created_at | datetime | YES | | NULL | | | updated_at | datetime | YES | | NULL | | +------------+----------+------+-----+---------+----------------+ Where is the "name" field? HELP ! Cheers !

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How to generate Doctrine models/classes that extend a custom record class

- by Shane O'Grady

When I use Doctrine to generate classes from Yaml/db each Base class (which includes the table definition) extends the Doctrine_Record class. Since my app uses a master and (multiple) slave db servers I need to be able to make the Base classes extend my custom record class to force writes to go to the master db server (as described here). However if I change the base class manually I lose it again when I regenerate my classes from Yaml/db using Doctrine. I need to find a way of telling Doctrine to extend my own Base class, or find a different solution to a master/slave db setup using Doctrine. Example generated model: abstract class My_Base_User extends Doctrine_Record { However I need it to be automatically generated as: abstract class My_Base_User extends My_Record { I am using Doctrine 1.2.1 in a new Zend Framework 1.9.6 application if it makes any difference.

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How do I find the "concrete class" of a django model baseclass

- by Mr Shark

I'm trying to find the actual class of a django-model object, when using model-inheritance. Some code to describe the problem: class Base(models.model): def basemethod(self): ... class Child_1(Base): pass class Child_2(Base): pass If I create various objects of the two Child classes and the create a queryset containing them all: Child_1().save() Child_2().save() (o1, o2) = Base.objects.all() I want to determine if the object is of type Child_1 or Child_2 in basemethod, I can get to the child object via o1.child_1 and o2.child_2 but that reconquers knowledge about the childclasses in the baseclass. I have come up with the following code: def concrete_instance(self): instance = None for subclass in self._meta.get_all_related_objects(): acc_name = subclass.get_accessor_name() try: instance = self.__getattribute__(acc_name) return instance except Exception, e: pass But it feels brittle and I'm not sure of what happens when if I inherit in more levels.

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Model objects versions in Django

- by pablo

Hi I'm building an e-commerce website. I have a Product and Order models. It's possible that a customer order a product and then the admin change its price or other fields before the customer actually get it. A possible solution is to add a 'version' field to the Product model. When the admin update a product field I'll add a timestamp and create a new object instead of updating the old one. An Order will have a reference to a specific product version. Does this make sense? Will overriding the Product Save method be sufficient to make it work? Thanks

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