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  • How do you determine subtype of an entity using Inheritance with Entity Framework 4?

    - by KallDrexx
    I am just starting to use the Entity Framework 4 for the first time ever. So far I am liking it but I am a bit confused on how to correctly do inheritance. I am doing a model-first approach, and I have my Person entity with two subtype entities, Employee and Client. EF is correctly using the table per type approach, however I can't seem to figure out how to determine what type of a Person a specific object is. For example, if I do something like var people = from p in entities.Person select p; return people.ToList<Person>(); In my list that I form from this, all I care about is the Id field so i don't want to actually query all the subtype tables (this is a webpage list with links, so all I need is the name and the Id, all in the Persons table). However, I want to form different lists using this one query, one for each type of person (so one list for Clients and another for Employees). The issue is if I have a Person entity, I can't see any way to determine if that entity is a Client or an Employee without querying the Client or Employee tables directly. How can I easily determine the subtype of an entity without performing a bunch of additional database queries?

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  • (Not So) Silly Objective-C inheritance problem when using property - GCC Bug?

    - by Ben Packard
    Update 2 - Many people are insisting I need to declare an iVar for the property. Some are saying not so, as I am using Modern Runtime (64 bit). I can confirm that I have been successfully using @property without iVars for months now. Therefore, I think the 'correct' answer is an explanation as to why on 64bit I suddenly have to explicitly declare the iVar when (and only when) i'm going to access it from a child class. The only one I've seen so far is a possible GCC bug (thanks Yuji). Not so simple after all... Update - I messed up one line of the original copy and paste - corrected. The @property call was missing (nonatomic, retain) but is a red herring - STILL NEED AN ANSWER! Thanks. I've been scratching my head with this for a couple of hours - I haven't used inheritance much. Here I have set up a simple Test B class that inherits from Test A, where an ivar is declared. But I get the compilation error that the variable is undeclared. This only happens when I add the property and synthesize declarations - works fine without them. TestA Header: #import <Cocoa/Cocoa.h> @interface TestA : NSObject { NSString *testString; } @end TestA Implementation is empty: #import "TestA.h" @implementation TestA @end TestB Header: #import <Cocoa/Cocoa.h> #import "TestA.h" @interface TestB : TestA { } @property (nonatomic, retain) NSString *testProp; @end TestB Implementation (Error - 'testString' is undeclared) #import "TestB.h" @implementation TestB @synthesize testProp; - (void)testing{ NSLog(@"test ivar is %@", testString); } @end

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  • How can I use a single-table inheritance and single controller to make this more DRY?

    - by Angela
    I have three models, Calls, Emails, and Letters and those are basically templates of what gets sent to individuals, modeled as Contacts. When a Call is made, a row in model in ContactCalls gets created. If an Email is sent, an entry in ContactEmails is made. Each has its own controller: contact_calls_controller.rb and contact_emails_controller.rb. I would like to create a single table inheritance called ContactEvents which has types Calls, Emails, and Letters. But I'm not clear how I pass the type information or how to consolidate the controllers. Here's the two controllers I have, as you can see, there's alot of duplication, but some differences that needs to be preserved. In the case of letter and postcards (another Model), it's even more so. class ContactEmailsController < ApplicationController def new @contact_email = ContactEmail.new @contact_email.contact_id = params[:contact] @contact_email.email_id = params[:email] @contact = Contact.find(params[:contact]) @company = Company.find(@contact.company_id) contacts = @company.contacts.collect(&:full_name) contacts.each do |contact| @colleagues = contacts.reject{ |c| [email protected]_name } end @email = Email.find(@contact_email.email_id) @contact_email.subject = @email.subject @contact_email.body = @email.message @email.message.gsub!("{FirstName}", @contact.first_name) @email.message.gsub!("{Company}", @contact.company_name) @email.message.gsub!("{Colleagues}", @colleagues.to_sentence) @email.message.gsub!("{NextWeek}", (Date.today + 7.days).strftime("%A, %B %d")) @contact_email.status = "sent" end def create @contact_email = ContactEmail.new(params[:contact_email]) @contact = Contact.find_by_id(@contact_email.contact_id) @email = Email.find_by_id(@contact_email.email_id) if @contact_email.save flash[:notice] = "Successfully created contact email." # send email using class in outbound_mailer.rb OutboundMailer.deliver_campaign_email(@contact,@contact_email) redirect_to todo_url else render :action => 'new' end end AND: class ContactCallsController < ApplicationController def new @contact_call = ContactCall.new @contact_call.contact_id = params[:contact] @contact_call.call_id = params[:call] @contact_call.status = params[:status] @contact = Contact.find(params[:contact]) @company = Company.find(@contact.company_id) @contact = Contact.find(@contact_call.contact_id) @call = Call.find(@contact_call.call_id) @contact_call.title = @call.title contacts = @company.contacts.collect(&:full_name) contacts.each do |contact| @colleagues = contacts.reject{ |c| [email protected]_name } end @contact_call.script = @call.script @call.script.gsub!("{FirstName}", @contact.first_name) @call.script.gsub!("{Company}", @contact.company_name ) @call.script.gsub!("{Colleagues}", @colleagues.to_sentence) end def create @contact_call = ContactCall.new(params[:contact_call]) if @contact_call.save flash[:notice] = "Successfully created contact call." redirect_to contact_path(@contact_call.contact_id) else render :action => 'new' end end

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  • How can I solve this CSS links inheritance problem?

    - by Craig Whitley
    It's stumped me an I've tried a couple of things - then again I'm not very experienced so I may just be going about it the wrong way. Basically I want to have different link styles for both the navigation and the pagination. The #navigation styling is overriding my .pagination styling though, and it doesn't appear to matter if the pagination is a class or an ID. I've also tried putting !important in the pagination styling, but this then makes the navigation inherit the pagination (been using firebug to check the inheritance). #navigation a:active, a:link, a:visited, a, a:focus { color: #ffde2f; font-family: Verdana, Geneva, Arial, Helvetica, sans-serif; font-size: 24px; text-decoration: none; } #navigation a:hover { color: #ffffff; font-family: Verdana, Geneva, Arial, Helvetica, sans-serif; font-size: 24px; text-decoration: none; } .pagination a:active, a:link, a:visited, a, a:focus { color: #fff; font-family: Verdana, Geneva, Arial, Helvetica, sans-serif; font-size: 14px; text-decoration: none; } .pagination { color: #fff; font-size: 14px; font-family: Verdana, Geneva, Arial, Helvetica, sans-serif; }

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  • Javascript inheritance: call super-constructor or use prototype chain?

    - by Jeremy S.
    Hi folks, quite recently I read about javascript call usage in MDC https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/Function/call one linke of the example shown below, I still don't understand. Why are they using inheritance here like this Prod_dept.prototype = new Product(); is this necessary? Because there is a call to the super-constructor in Prod_dept() anyway, like this Product.call is this just out of common behaviour? When is it better to use call for the super-constructor or use the prototype chain? function Product(name, value){ this.name = name; if(value >= 1000) this.value = 999; else this.value = value; } function Prod_dept(name, value, dept){ this.dept = dept; Product.call(this, name, value); } Prod_dept.prototype = new Product(); // since 5 is less than 1000, value is set cheese = new Prod_dept("feta", 5, "food"); // since 5000 is above 1000, value will be 999 car = new Prod_dept("honda", 5000, "auto"); Thanks for making things clearer

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  • How can I achieve this kind of relationship (inheritance, composition, something else)?

    - by Tim
    I would like to set up a foundation of classes for an application, two of which are person and student. A person may or may not be a student and a student is always a person. The fact that a student “is a” person led me to try inheritance, but I can't see how to make it work in the case where I have a DAO that returns an instance of person and I then want to determine if that person is a student and call student related methods for it. class Person { private $_firstName; public function isStudent() { // figure out if this person is a student return true; // (or false) } } class Student extends Person { private $_gpa; public function getGpa() { // do something to retrieve this student's gpa return 4.0; // (or whatever it is) } } class SomeDaoThatReturnsPersonInstances { public function find() { return new Person(); } } $myPerson = SomeDaoThatReturnsPersonInstances::find(); if($myPerson->isStudent()) { echo 'My person\'s GPA is: ', $myPerson->getGpa(); } This obviously doesn't work, but what is the best way to achieve this effect? Composition doesn't sond right in my mind because a person does not “have a” student. I'm not looking for a solution necessarily but maybe just a term or phrase to search for. Since I'm not really sure what to call what I'm trying to do, I haven't had much luck. Thank you!

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  • Generic relations missing with grappelli

    - by diegueus9
    I'm using the last svn revision of grappelli and rev 11840 of django (before multidatabases and stuff), and i'm trying to use generic relations in the admin, but doesn't work, The model: class AutorProyectoLey(DatedModel): tipo_autor = models.ForeignKey(ContentType) autor_id = models.PositiveIntegerField() content_object = generic.GenericForeignKey('tipo_autor', 'autor_id') proyecto_ley = models.ForeignKey(ProyectoLey) The admin: class AutorInline(GenericInlineModelAdmin): model = AutorProyectoLey allow_add = True ct_field = 'tipo_autor' ct_fk_field = 'autor_id' classes = ('collapse-open',) And i put this model of var inlines in another adminmodel, but the html render is : <!-- Inlines --> <!-- Submit-Row -->

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  • 2 fields unique

    - by xRobot
    I have this model: class blog(models.Model): user = models.ForeignKey(User) mail = models.EmailField(max_length=60, null=False, blank=False) name = models.CharField(max_length=60, blank=True, null=True) I want that (user,email) are unique togheter. For example: This is allowed: 1, [email protected], myblog 2, [email protected], secondblog This is NOT allowed: 1, [email protected], myblog 1, [email protected], secondblog Is this possible in Django ?

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  • why save_model method doesn't work in admin.StackedInline?

    - by FurtiveFelon
    Hi all, I have a similar problem as a previously solved problem of mine, except this time solution doesn't seem to work: http://stackoverflow.com/questions/2991365/how-to-auto-insert-the-current-user-when-creating-an-object-in-django-admin Previously i used to override the save_model to stamp the user submitting the article. Now i need to do the same for comments, it doesn't seem to work anymore. Anyone have any ideas? Thanks a lot! Jason

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  • How to use the same template for different query sets?

    - by knuckfubuck
    I'm new to Django and setting up my first site. I have a Share model and a template called share_list.html that uses an object_list like this: {% for object in object_list %} I setup haystack using their tutorial and the search template looks like this: {% for result in page.object_list %} I would like to modify the search.html template to have an include of the share_list so I don't have to repeat myself. How can I make it use the same object_list?

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  • apache solr auto suggestions

    - by Pydev UA
    I use solr+django-haystack I set settings.HAYSTACK_INCLUDE_SPELLING = True and rebuild index I'm trying to get any suggestion using: SearchQuerySet().auto_query('tryng ani word her').spelling_suggestion() But I always get None What should I do to get at least one working suggestion ? may be I need add some configuration into solr config or have some specific data indexed ?

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  • regular expression with special chars

    - by xRobot
    I need a regular expression to validate string with one or more of these characters: a-z A-Z ' àòèéùì simple white space FOR EXAMPLE these string are valide: D' argon calabrò maryòn l' Ancol these string are NOT valide: hello38239 my_house work [tab] with me I tryed this: re.match(r"^[a-zA-Z 'òàèéìù]+$", self.cleaned_data['title'].strip()) It seems to work in my python shell but in Django I get this error: SyntaxError at /home/ ("Non-ASCII character '\\xc3' ... Why ?

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  • Overriding initial value in ModelForm

    - by schneck
    Hi, in my Django (1.2) project, I want to prepopulate a field in a modelform, but my new value is ignored. This is the snippet: class ArtefactForm(ModelForm): material = CharField(widget=AutoCompleteWidget('material', force_selection=False)) def __init__(self, *args, **kwargs): super(ArtefactForm, self).__init__(*args, **kwargs) self.fields['material'].initial = 'Test' I also tried with self.base_fields, but no effect: there is always the database-value displaying in the form. Any ideas?

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  • PHP Frameworks (CodeIgnitor, Yii, CakePHP) vs. Django

    - by niting
    I have to develop a site which has to accomodate around 2000 users a day and speed is a criterion for it. Moreover, the site is a user oriented one where the user will be able to log in and check his profile, register for specific events he/she wants to participate in. The site is to be hosted on a VPS server.Although I have pretty good experience with python and PHP but I have no idea how to use either of the framework. We have plenty of time to experiment and learn one of the above frameworks.Could you please specify which one would be preferred for such a scenario considering speed, features, and security of the site. Thanks, niting

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  • List display names from django models

    - by Ed
    I have an object: POP_CULTURE_TYPES = ( ('SG','Song'), ('MV', 'Movie'), ('GM', 'Game'), ('TV', 'TV'), ) class Pop_Culture(models.Model): name = models.CharField(max_length=30, unique=True) type = models.CharField(max_length=2, choices = POP_CULTURE_TYPES, blank=True, null=True) Then I have a function: def choice_list(request, modelname, field_name): mdlnm = get.model('mdb', modelname.lower()) mdlnm = mdlnm.objects.values_list(field_name, flat=True).distinct().order_by(field_name) return render_to_response("choice_list.html", { 'model' : modelname, 'field' : field_name, 'field_list' : mdlnm }) This gives me a distinct list of all the "type" entries in the database in the "field_list" variable passed in render_to_response. But I don't want a list that shows: SG MV I want a list that shows: Song Movie I can do this on an individual object basis if I was in the template object.get_type_display But how do I get a list of all of the unique "type" entries in the database as their full names for output into a template? I hope this question was clearly described. . .

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  • Validating ModelChoiceField in Django forms

    - by Andrey
    I'm trying to validate a form containing a ModelChoiceField: state = forms.ModelChoiceField(queryset=State.objects.all(), empty_label=None) When it is used in normal circumstances, everything goes just fine. But I'd like to protect the form from the invalid input. It's pretty obvious that I must get forms.ValidationError when I put invalid value in this field, isn't it? But if I try to submit a form with a value 'invalid' in 'state' field, I get ValueError: invalid literal for int() with base 10: 'invalid' and not the expected forms.ValidationError. What should I do? I tried to place a def clean_state(self) to check this field but that didn't work plus I don't think this is a good solution, there must be something more simple but I just missed that.

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  • Generate unique hashes for django models

    - by becomingGuru
    I want to use unique hashes for each model rather than ids. I implemented the following function to use it across the board easily. import random,hashlib from base64 import urlsafe_b64encode def set_unique_random_value(model_object,field_name='hash_uuid',length=5,use_sha=True,urlencode=False): while 1: uuid_number = str(random.random())[2:] uuid = hashlib.sha256(uuid_number).hexdigest() if use_sha else uuid_number uuid = uuid[:length] if urlencode: uuid = urlsafe_b64encode(uuid)[:-1] hash_id_dict = {field_name:uuid} try: model_object.__class__.objects.get(**hash_id_dict) except model_object.__class__.DoesNotExist: setattr(model_object,field_name,uuid) return I'm seeking feedback, how else could I do it? How can I improve it? What is good bad and ugly about it?

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  • GAE/Django Templates (0.96) filters to get LENGTH of GqlQuery and filter it

    - by Halst
    I pass the query with comments to my template: COMM = CommentModel.gql("ORDER BY created") doRender(self,CP.template,{'CP':CP,'COMM':COMM, 'authorize':authorize()}) And I want to output the number of comments as a result, and I try to do things like that: <a href="...">{{ COMM|length }} comments</a> Thats does not work (yeah, since COMM is GqlQuery, not a list). What can I do with that? Is there a way to convert GqlQuery to list or is there another solution? (first question) Second question is, how to filter this list in template? Is there a construct like this: <a href="...">{{ COMM|where(reference=smth)|length }} comments</a> so that I can get not only the number of all comments, but only comments with certain db.ReferenceProperty() property, for example. Last question: is it weird to do such things using templates?

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  • Python Django MySQLdb setup problem:: setup.py dosen't build due to incorrect location of mysql

    - by 108860375137931889948
    I'm trying to install MySQLdb for python. but when I run the setup, this is the error I get. well I know why its giving all the missing file statements, but dont know where to change the bold marked location from. Please help gaurav-toshniwals-macbook-7:MySQL-python-1.2.3c1 gauravtoshniwal$ python setup.py build running build running build_py copying MySQLdb/release.py - build/lib.macosx-10.3-fat-2.6/MySQLdb running build_ext building '_mysql' extension gcc-4.0 -arch ppc -arch i386 -isysroot /Developer/SDKs/MacOSX10.4u.sdk -fno-strict-aliasing -fno-common -dynamic -DNDEBUG -g -O3 -Dversion_info=(1,2,3,'gamma',1) -D_version_=1.2.3c1 -I/Applications/MAMP/Library/include/mysql -I/Library/Frameworks/Python.framework/Versions/2.6/include/python2.6 -c _mysql.c -o build/temp.macosx-10.3-fat-2.6/_mysql.o _mysql.c:36:23: error: my_config.h: No such file or directory _mysql.c:36:23: error: my_config.h: No such file or directory _mysql.c:38:19: error: mysql.h: No such file or directory _mysql.c:38:19:_mysql.c:39:26: error: mysqld_error.h: No such file or directory error: _mysql.c:40:20:mysql.h: No such file or directory

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  • How to manage Javascript modules in django templates?

    - by John Mee
    Lets say we want a library of javascript-based pieces of functionality (I'm thinking jquery): For example: an ajax dialog a date picker a form validator a sliding menu bar an accordian thingy There are four pieces of code for each: some Python, CSS, JS, & HTML. What is the best way to arrange all these pieces so that: each javascript 'module' can be neatly reused by different views the four bits of code that make up the completed function stay together the css/js/html parts appear in their correct places in the response common dependencies between modules are not repeated (eg: a javascript file in common) x-------------- It would be nice if, or is there some way to ensure that, when called from a templatetag, the templates respected the {% block %} directives. Thus one could create a single template with a block each for CSS, HTML, and JS, in a single file. Invoke that via a templatetag which is called from the template of whichever view wants it. That make any sense. Can that be done some way already? My templatetag templates seem to ignore the {% block %} directives. x-------------- There's some very relevant gasbagging about putting such media in forms here http://docs.djangoproject.com/en/dev/topics/forms/media/ which probably apply to the form validator and date picker examples.

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  • ContentType Issue -- Human is an idiot - Can't figure out how to tie the original model to a Content

    - by bmelton
    Originally started here: http://stackoverflow.com/questions/2650181/django-in-query-as-a-string-result-invalid-literal-for-int-with-base-10 I have a number of apps within my site, currently working with a simple "Blog" app. I have developed a 'Favorite' app, easily enough, that leverages the ContentType framework in Django to allow me to have a 'favorite' of any type... trying to go the other way, however, I don't know what I'm doing, and can't find any examples for. I'll start off with the favorite model: favorite/models.py from django.db import models from django.contrib.contenttypes.models import ContentType from django.contrib.contenttypes import generic from django.contrib.auth.models import User class Favorite(models.Model): content_type = models.ForeignKey(ContentType) object_id = models.PositiveIntegerField() user = models.ForeignKey(User) content_object = generic.GenericForeignKey() class Admin: list_display = ('key', 'id', 'user') class Meta: unique_together = ("content_type", "object_id", "user") Now, that allows me to loop through the favorites (on a user's "favorites" page, for example) and get the associated blog objects via {{ favorite.content_object.title }}. What I want now, and can't figure out, is what I need to do to the blog model to allow me to have some tether to the favorite (so when it is displayed in a list it can be highlighted, for example). Here is the blog model: blog/models.py from django.db import models from django.db.models import permalink from django.template.defaultfilters import slugify from category.models import Category from section.models import Section from favorite.models import Favorite from django.contrib.auth.models import User from django.contrib.contenttypes.models import ContentType from django.contrib.contenttypes import generic class Blog(models.Model): title = models.CharField(max_length=200, unique=True) slug = models.SlugField(max_length=140, editable=False) author = models.ForeignKey(User) homepage = models.URLField() feed = models.URLField() description = models.TextField() page_views = models.IntegerField(null=True, blank=True, default=0 ) created_on = models.DateTimeField(auto_now_add = True) updated_on = models.DateTimeField(auto_now = True) def __unicode__(self): return self.title @models.permalink def get_absolute_url(self): return ('blog.views.show', [str(self.slug)]) def save(self, *args, **kwargs): if not self.slug: slug = slugify(self.title) duplicate_count = Blog.objects.filter(slug__startswith = slug).count() if duplicate_count: slug = slug + str(duplicate_count) self.slug = slug super(Blog, self).save(*args, **kwargs) class Entry(models.Model): blog = models.ForeignKey('Blog') title = models.CharField(max_length=200) slug = models.SlugField(max_length=140, editable=False) description = models.TextField() url = models.URLField(unique=True) image = models.URLField(blank=True, null=True) created_on = models.DateTimeField(auto_now_add = True) def __unicode__(self): return self.title def save(self, *args, **kwargs): if not self.slug: slug = slugify(self.title) duplicate_count = Entry.objects.filter(slug__startswith = slug).count() if duplicate_count: slug = slug + str(duplicate_count) self.slug = slug super(Entry, self).save(*args, **kwargs) class Meta: verbose_name = "Entry" verbose_name_plural = "Entries" Any guidance?

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