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  • Django: order by count of a ForeignKey field?

    - by AP257
    This is almost certainly a duplicate question, in which case apologies, but I've been searching for around half an hour on SO and can't find the answer here. I'm probably using the wrong search terms, sorry. I have a User model and a Submission model. Each Submission has a ForeignKey field called user_submitted for the User who uploaded it. class Submission(models.Model): uploaded_by = models.ForeignKey('User') class User(models.Model): name = models.CharField(max_length=250 ) My question is pretty simple: how can I get a list of the three users with the most Submissions? I trued creating a num_submissions method on the User model: def num_submissions(self): num_submissions = Submission.objects.filter(uploaded_by=self).count() return num_submissions and then doing: top_users = User.objects.filter(problem_user=False).order_by('num_submissions')[:3] but this fails, as do all the other things I've tried. Can I actually do it using a smart database query? Or should I just do something more hacky in the views file?

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  • In partial view: "The model item passed into the dictionary is of type"

    - by Dave
    I lack understanding of some basic MVC concepts, despite all my searching. I created an MVC project in Visual Studio, which contains the partial view _LogOnPartial.shtml. I just want to access information within the view pertaining to the user, to put in a user dropdown menu. When I try to put this at the top of the partial view cshtml page I get the above error: @model MyProject_MVC.Models.UserRepository When I try this I also get an error: @Html.Partial("_LogOnPartial", MyProject_MVC.Models.UserRepository) 'MyProject_MVC.Models.UserRepository' is a 'type', which is not valid in the given context

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  • Python Importing object that originates in one module from a different module into a third module

    - by adewinter
    I was reading the sourcode for a python project and came across the following line: from couchexport.export import Format (source: https://github.com/wbnigeria/couchexport/blob/master/couchexport/views.py#L1 ) I went over to couchexport/export.py to see what Format was (Class? Dict? something else?). Unfortunately Format isn't in that file. export.py does however import a Format from couchexport.models where there is a Format class (source: https://github.com/wbnigeria/couchexport/blob/master/couchexport/models.py#L11). When I open up the original file in my IDE and have it look up the declaration, in line I mentioned at the start of this question, it leads directly to models.py. What's going on? How can an import from one file (export.py) actually be an import from another file (models.py) without being explicitly stated?

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  • Single database with multiple instances of Django

    - by jwesonga
    I have a Django project where the company will have a main site like www.ourcompany.org and a bunch of sub-domains like project.ourcompany.org. Content appearing in the sub-domains like case studies should also appear in the main site. I've decided to use multiple instances of Django BUT one database for each sub-domain so that I can have some flexibility and take advantage of the Sites framework. What I'm not sure of is how to access the models across the multiple instances. If I have a model: class CaseStudy(models.Model): title=models.CharField(max_length=100) site=models.ManyToMany(Site) Do I need to create this model in every instance so that I can have access to the object?

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  • Ruby-on-Rails equivalent to ORM Designer for Symfony?

    - by fayer
    In Symfony i just have to create models with ORM Designer and export it to symfony as a schema.yml and then use a symfony command to create tables, models and forms. I wonder if there is an equivalent to the RoR so that you dont have to create models manually by hand? It saves a lot of time using GUI for this kind of tasks and it is less error-prone. thanks

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  • Django: How to handle imports in a reusable app

    - by facha
    everyone I'm just starting with django. It is not quite clear to me, how should I write an app I could reuse later. In every tutorial I read I see the same piece of code: view.py from project.app.models import MyModel So, if I move my apps to another project, I'll have to modify the "project.app.models" so that it looks like "project2.app.models" for every app I move. Is there a way to avoid that? Thanks in advance.

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  • Rails 3 ActiveModel Nested Class I18n

    - by Dave
    Given the following class definition in ruby: class Conversation class Message include ActiveModel::Validations attr_accessor :quantity validates :quantity, :presence => true end end How can you use i18n to customize to error message. For example the correct lookup for the class Conversation would be activemodel: errors: models: conversation: attributes: quantity: blank: "Some custom message" But what is it for the Message class? I tried: activemodel: errors: models: conversation: message: attributes: quantity: blank: "Some custom message" activemodel: errors: models: message: attributes: quantity: blank: "Some custom message" activemodel: errors: models: conversation::message: attributes: quantity: blank: "Some custom message" None of them work Any ideas or is this a bug with ActiveModel or I18n?

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  • current_user.user_type_id = @employer ID

    - by sscirrus
    I am building a system with a User model (authenticated using AuthLogic) and three user types in three models: one of these models is Employer. Each of these three models has_many :users, :as = :authenticable. I start by having a new visitor to the site create their own 'User' record with username, password, which user type they are, etc. Upon creation, the user is sent to the 'new' action for one of the three models. So, if they tell us they are an employer, we redirect_to :controller = "employers, :action = "new". Question: When the employer has submitted, I want to set the current_user.user_type_id equal to the employer ID. This should be simple... but it's not working. # Employers Controller / new def new @employer = Employer.new 1.times {@employer.addresses.build} render :layout => 'forms' end # Employers Controller / create def create @employer = Employer.new(params[:employer]) if @employer.save if current_user.blank? redirect_to :controller => "users", :action => "new" else current_user.user_type_id = @employer.id current_user.user_type = "Employer" redirect_to :action => "home", :id => current_user.user_type_id end else render :action => "new" end end ------UPDATE------ Hi guys. In response: I am using this table structure because each of my three user type models have lots of different fields and each has different relationships to the other models, which is why I've avoided STI. By 1.times (@employer.addresses.build) I'm connecting the employer model to the address polymorphic table in one form, so I'm asking the controller to build a new address to go along with the new employer. Averell: you mentioned encapsulating... something in the model using a 'setter' method. I have no idea what you mean by this - could you please explain how this works (or direct me to an example elsewhere)? With tsdbrown's answer I have managed to create the behavior I want... if there's a more elegant way to accomplish the same thing I'd love to learn how. Thanks very much. Thanks to tsdbrown for answering the current_user.save problem!

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  • Django Model Formset Pre-Filled Value Problem

    - by user552377
    Hi, i'm trying to use model formsets with Django. When i load forms template, i see that it's filled-up with previous values. Is there a caching mechanism that i should stop, or what? Thanks for your help, here is my code: models.py class FooModel( models.Model ): a_field = models.FloatField() b_field = models.FloatField() def __unicode__( self ): return self.a_field forms.py from django.forms.models import modelformset_factory FooFormSet = modelformset_factory(FooModel) views.py def foo_func(request): if request.method == 'POST': formset = FooFormSet(request.POST, request.FILES, prefix='foo_prefix' ) if formset.is_valid(): formset.save() return HttpResponseRedirect( '/true/' ) else: return HttpResponseRedirect( '/false/' ) else: formset = FooFormSet(prefix='foo_prefix') variables = RequestContext( request , { 'formset':formset , } ) return render_to_response('footemplate.html' , variables ) template: <form method="post" action="."> {% csrf_token %} <input type="submit" value="Submit" /> <table id="FormsetTable" border="0" cellpadding="0" cellspacing="0"> <tbody> {% for form in formset.forms %} <tr> <td>{{ form.a_field }}</td> <td>{{ form.b_field }}</td> </tr> {% endfor %} </tbody> </table> {{ formset.management_form }} </form>

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  • Django - partially validating form

    - by aeter
    I'm new to Django, trying to process some forms. I have this form for entering information (creating a new ad) in one template: class Ad(models.Model): ... category = models.CharField("Category",max_length=30, choices=CATEGORIES) sub_category = models.CharField("Subcategory",max_length=4, choices=SUBCATEGORIES) location = models.CharField("Location",max_length=30, blank=True) title = models.CharField("Title",max_length=50) ... I validate it with "is_valid()" just fine. Basically for the second validation (another template) I want to validate only against "category" and "sub_category": In another template, I want to use 2 fields from the same form ("category" and "sub_category") for filtering information - and now the "is_valid()" method would not work correctly, cause it validates the entire form, and I need to validate only 2 fields. I have tried with the following: ... if request.method == 'POST': # If a filter for data has been submitted: form = AdForm(request.POST) try: form = form.clean() category = form.category sub_category = form.sub_category latest_ads_list = Ad.objects.filter(category=category) except ValidationError: latest_ads_list = Ad.objects.all().order_by('pub_date') else: latest_ads_list = Ad.objects.all().order_by('pub_date') form = AdForm() ... but it doesn't work. How can I validate only the 2 fields category and sub_category?

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  • Django ModelForm Imagefield Upload

    - by Wei Xu
    I am pretty new to Django and I met a problem in handling image upload using ModelForm. My model is as following: class Project(models.Model): name = models.CharField(max_length=100) description = models.CharField(max_length=2000) startDate = models.DateField(auto_now_add=True) photo = models.ImageField(upload_to="projectimg/", null=True, blank=True) And the modelform is as following: class AddProjectForm(ModelForm): class Meta: model = Project widgets = { 'description': Textarea(attrs={'cols': 80, 'rows': 50}), } fields = ['name', 'description', 'photo'] And the View function is: def addProject(request, template_name): if request.method == 'POST': addprojectform = AddProjectForm(request.POST,request.FILES) print addprojectform if addprojectform.is_valid(): newproject = addprojectform.save(commit=False) print newproject print request.FILES newproject.photo = request.FILES['photo'] newproject.save() print newproject.photo else: addprojectform = AddProjectForm() newProposalNum = projectProposal.objects.filter(solved=False).count() return render(request, template_name, {'addprojectform':addprojectform, 'newProposalNum':newProposalNum}) the template is: <form class="bs-example form-horizontal" method="post" action="">{% csrf_token %} <h2>Project Name</h2><br> {{ addprojectform.name }}<br> <h2>Project Description</h2> {{ addprojectform.description }}<br> <h2>Image Upload</h2><br> {{ addprojectform.photo }}<br> <input type="submit" class="btn btn-success" value="Add Project"> </form> Can anyone help me or could you give an example of image uploading? Thank you!

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  • LINQ EF not saving to database...

    - by Keith Barrows
    I guess this is a continuation of the last question I asked: http://stackoverflow.com/questions/2587542/bulk-insert-and-update-with-ado-net-entity-framework. I am not getting any errors while doing inserts yet no data is actually going into my DB. My DB is a SDF file (SQL CE). Any ideas what to check? My app.config looks like: <?xml version="1.0" encoding="utf-8"?> <configuration> <configSections> </configSections> <connectionStrings> <add name="Lab_Use_Billing.Properties.Settings.LabUseConnectionString" connectionString="Data Source=|DataDirectory|\Models\LabUse.sdf" providerName="Microsoft.SqlServerCe.Client.3.5" /> <add name="LabUseEntities" connectionString="metadata=res://*/Models.LabUseEntities.csdl|res://*/Models.LabUseEntities.ssdl|res://*/Models.LabUseEntities.msl; provider=System.Data.SqlServerCe.3.5; provider connection string=&quot;Data Source=|DataDirectory|\Models\LabUse.sdf&quot;" providerName="System.Data.EntityClient" /> </connectionStrings> </configuration> TIA

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  • Django admin - remove field if editing an object

    - by John McCollum
    I have a model which is accessible through the Django admin area, something like the following: # model class Foo(models.Model): field_a = models.CharField(max_length=100) field_b = models.CharField(max_length=100) # admin.py class FooAdmin(admin.ModelAdmin): pass Let's say that I want to show field_a and field_b if the user is adding an object, but only field_a if the user is editing an object. Is there a simple way to do this, perhaps using the fields attribute? If if comes to it, I could hack a JavaScript solution, but it doesn't feel right to do that at all!

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  • MVC 3 Nested EditorFor Templates

    - by Gordon Hickley
    I am working with MVC 3, Razor views and EditorFor templates. I have three simple nested models:- public class BillingMatrixViewModel { public ICollection<BillingRateRowViewModel> BillingRateRows { get; set; } public BillingMatrixViewModel() { BillingRateRows = new Collection<BillingRateRowViewModel>(); } } public class BillingRateRowViewModel { public ICollection<BillingRate> BillingRates { get; set; } public BillingRateRowViewModel() { BillingRates = new Collection<BillingRate>(); } } public class BillingRate { public int Id { get; set; } public int Rate { get; set; } } The BillingMatrixViewModel has a view:- @using System.Collections @using WIP_Data_Migration.Models.ViewModels @model WIP_Data_Migration.Models.ViewModels.BillingMatrixViewModel <table class="matrix" id="matrix"> <tbody> <tr> @Html.EditorFor(model => Model.BillingRateRows, "BillingRateRow") </tr> </tbody> </table> The BillingRateRow has an Editor Template called BillingRateRow:- @using System.Collections @model IEnumerable<WIP_Data_Migration.Models.ViewModels.BillingRateRowViewModel> @foreach (var item in Model) { <tr> <td> @item.BillingRates.First().LabourClass.Name </td> @Html.EditorFor(m => item.BillingRates) </tr> } The BillingRate has an Editor Template:- @model WIP_Data_Migration.Models.BillingRate <td> @Html.TextBoxFor(model => model.Rate, new {style = "width: 20px"}) </td> The markup produced for each input is:- <input name="BillingMatrix.BillingRateRows.item.BillingRates[0].Rate" id="BillingMatrix_BillingRateRows_item_BillingRates_0__Rate" style="width: 20px;" type="text" value="0"/> Notice the name and ID attributes the BillingRate indexes are handled nicely but the BillingRateRows has no index instead '.item.'. From my reasearch this is because the context has been pulled out due to the foreach loop, the loop shouldn't be necessary. I want to achieve:- <input name="BillingMatrix.BillingRateRows[0].BillingRates[0].Rate" id="BillingMatrix_BillingRateRows_0_BillingRates_0__Rate" style="width: 20px;" type="text" value="0"/> If I change the BillingRateRow View to:- @model WIP_Data_Migration.Models.ViewModels.BillingRateRowViewModel <tr> @Html.EditorFor(m => Model.BillingRates) </tr> It will throw an InvalidOperationException, 'model item passed into the dictionary is of type System.Collections.ObjectModel.Collection [BillingRateRowViewModel] but this dictionary required a type of BillingRateRowViewModel. Can anyone shed any light on this?

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  • Getting a random element in Django

    - by Sarah
    I just finished the Django tutorial and started work on my own project, however, I seem to have missed something completely. I wanted to get a random slogan from this model: from django.db import models class Slogan(models.Model): slogan = models.CharField(max_length=200) And return it in this view: from django.http import HttpResponse from swarm.sloganrotator.models import Slogan def index(request): return HttpResponse(Slogan.objects.order_by('?')[:1]) However, the view just returns 'Slogan object'. Then I thought, maybe I can access the slogan string itself by simply appending .slogan to the slice, but that gives me an error indicating that the object I have is actually a QuerySet and has no attribute slogan. I've obviously misunderstood something about Django here, but it just doesn't fall into place for me. Any help?

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  • Design question?

    - by Mohamed
    I am building music app, where user can do several tasks including but not limited to listening song, like song, recommend song to a friend and extra. currently I have this model: class Activity(models.Model): activity = models.TextField() user = models.ForeignKey(User) date = models.DateTimeField(auto_now=True) so far I thought about two solutions. 1. saving a string to database. e.g "you listened song xyz" 2. create a dictionary about the activity and save to the database using pickle or json. e.g. dict_ = {"activity_type":"listening", "song":song_obj} I am leaning to the second implementation, but not quite sure. so what do you think about those two methods? do you know better way to achieve the goal?

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  • Sorting related objects in the Django Admin form interface

    - by Carver
    I am looking to sort the related objects that show up when editing an object using the admin form. So for example, I would like to take the following object: class Person(models.Model): first_name = models.CharField( ... ) last_name = models.CharField( ... ) hero = models.ForeignKey( 'self', null=True, blank=True ) and edit the first name, last name and hero using the admin interface. I want to sort the objects as they show up in the drop down by last name, first name (ascending). How do I do that? Context I'm using Django v1.1. I started by looking for help in the django admin docs, but didn't find the solution As you can see in the example, the foreign key is pointing to itself, but I expect it would be the same as pointing to a different model object. Bonus points for being able to filter the related objects, too (eg~ only allow selecting a hero with the same first name)

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  • Need help optimizing this Django aggregate query

    - by Chris Lawlor
    I have the following model class Plugin(models.Model): name = models.CharField(max_length=50) # more fields which represents a plugin that can be downloaded from my site. To track downloads, I have class Download(models.Model): plugin = models.ForiegnKey(Plugin) timestamp = models.DateTimeField(auto_now=True) So to build a view showing plugins sorted by downloads, I have the following query: # pbd is plugins by download - commented here to prevent scrolling pbd = Plugin.objects.annotate(dl_total=Count('download')).order_by('-dl_total') Which works, but is very slow. With only 1,000 plugins, the avg. response is 3.6 - 3.9 seconds (devserver with local PostgreSQL db), where a similar view with a much simpler query (sorting by plugin release date) takes 160 ms or so. I'm looking for suggestions on how to optimize this query. I'd really prefer that the query return Plugin objects (as opposed to using values) since I'm sharing the same template for the other views (Plugins by rating, Plugins by release date, etc.), so the template is expecting Plugin objects - plus I'm not sure how I would get things like the absolute_url without a reference to the plugin object. Or, is my whole approach doomed to failure? Is there a better way to track downloads? I ultimately want to provide users some nice download statistics for the plugins they've uploaded - like downloads per day/week/month. Will I have to calculate and cache Downloads at some point? EDIT: In my test dataset, there are somewhere between 10-20 Download instances per Plugin - in production I expect this number would be much higher for many of the plugins.

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  • Django - Expression based model constraints

    - by rtmie
    Is it possible to set an expression based constraint on a django model object, e.g. If I want to impose a constraint where an owner can have only one widget of a given type that is not in an expired state, but can have as many others as long as they are expired. Obviously I can do this by overriding the save method, but I am wondering if it can be done by setting constraints, e.g. some derivative of the unique_together constraint WIDGET_STATE_CHOICES = ( ('NEW', 'NEW'), ('ACTIVE', 'ACTIVE'), ('EXPIRED', 'EXPIRED') ) class MyWidget(models.Model): owner = models.CharField(max_length=64) widget_type = models.CharField(max_length = 10) widget_state = models.CharField(max_length = 10, choices = WIDGET_STATE_CHOICES) #I'd like to be able to do something like class Meta: unique_together = (("owner","widget_type","widget_state" != 'EXPIRED')

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  • filter queryset based on list, including None

    - by jujule
    Hi all I dont know if its a django bug or a feature but i have a strange ORM behaviour with MySQL. class Status(models.Model): name = models.CharField(max_length = 50) class Article(models.Model) status = models.ForeignKey(status, blank = True, null=True) filters = Q(status__in =[0, 1,2] ) | Q(status=None) items = Article.objects.filter(filters) this returns Article items but some have other status than requested [0,1,2,None] looking at the sql query : SELECT [..] FROM `app_article` LEFT OUTER JOIN `app_status` ON (`app_article`.`status_id` = `app_status`.`id`) WHERE (`app_article`.`status_id` IN (1, 2) OR `app_status`.`id` IS NULL) ORDER BY [...] the OR app_status.id IS NULL part seems to be the cause. if i change it to OR app_article.status_id IS NULL it works correctly. How to deal with this ? Thanx.

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  • inheritance from the django user model results in error when changing password

    - by Jerome
    I inherited form the django user model like so: from django.db import models from django.contrib.auth.models import User, UserManager from django.utils.translation import ugettext_lazy as _ class NewUserModel(User): custom_field_1 = models.CharField(_('custom field 1'), max_length=250, null=True, blank=True) custom_field_2 = models.CharField(_('custom field 2'), max_length=250, null=True, blank=True) objects = UserManager() When i go to the admin and add an entry into this model, it saves fine, but below the "Password" field where it has this text "Use '[algo]$[salt]$[hexdigest]' or use the change password form.", if i click on the "change password form' link, it produces this error Truncated incorrect DOUBLE value: '7/password' What can i do to fix this?

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  • best way to create tables with ORM?

    - by ajsie
    assume that i start coding an application from scratch, is the best way to create tables when using an ORM (doctrine), to manually create tables in mysql and then generate models from the tables, or is it the other way around, that is to create the models in php and then generate tables from models? and if i already have a database, will the models created be optimal? cause i have heard some say that its best to create the database from scratch when using ORM, so that the relations are optimized for OOD. share your thoughts!

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  • [Django] One single page to create a Parent object and its associated child objects

    - by ahmoo
    Hi all, This is my very first post on this awesome site, from which I have been finding answers to a handful of challenging questions. Kudos to the community! I am new to the Django world, so am hoping to find help from some Django experts here. Thanks in advance. Item model: class Item(models.Model): name = models.CharField(max_length=50) ItemImage model: class ItemImage(models.Model): image = models.ImageField(upload_to=get_unique_filename) item = models.ForeignKey(Item, related_name='images') As you can tell from the model definitions above, every Item object can have many ItemImage objects. My requirements are as followings: A single web page that allows users to create a new Item while uploading the images associated with the Item. The Item and the ItemImages objects should be created in the database all together, when the "Save" button on the page is clicked. I have created a variable in a custom config file, called NUMBER_OF_IMAGES_PER_ITEM. It is based on this variable that the system generates the number of image fields per item. Questions: What should the forms and the template be like? Can ModelForm be used to achieve the requirements? For the view function, what do I need to watch out other than making sure to save Item before ItemImage objects?

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