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  • Model Binding, a simple, simple question

    - by Paul Hatcherian
    I have a struct which works much like the System.Nullable type: public struct SpecialProperty<T> { public static implicit operator T(SpecialProperty<T> value) { return value.Value; } public static implicit operator SpecialProperty<T>(T value) { return new TrackChanges<T> { Value = value }; } T internalValue; public T Value { get { return internalValue; } set { internalValue = value; } } public override bool Equals(object other) { return Value.Equals(other); } public override int GetHashCode() { return Value.GetHashCode(); } public override string ToString() { return Value.ToString(); } } I'm trying to use it with ASP.NET MVC binding. Using the default customer model binder the property will always yield null. I can fix this by adding ".Value" to the end of every form input name, but I just want it to bind to the new type directly using some sort of custom model binder, but all the solutions I've tried seemed needlessly complex. I feel like I should be able to extend the default binder and with a few lines of code redirect the property binding to the entire model using implicit conversion. I don't quite get the binding paradigm of the default binder, but it seems really stuck on this distinction between the model and model properties. What is the simplest method to do this? Thanks!

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  • MultiWidget in MultiWidget how to compress the first one?

    - by sacabuche
    I have two MultiWidget one inside the other, but the problem is that the MultiWidget contained don't return compress, how do i do to get the right value from the first widget? In this case from SplitTimeWidget class SplitTimeWidget(forms.MultiWidget): """ Widget written to split widget into hours and minutes. """ def __init__(self, attrs=None): widgets = ( forms.Select(attrs=attrs, choices=([(hour,hour) for hour in range(0,24)])), forms.Select(attrs=attrs, choices=([(minute, str(minute).zfill(2)) for minute in range(0,60)])), ) super(SplitTimeWidget, self).__init__(widgets, attrs) def decompress(self, value): if value: return [value.hour, value.minute] return [None, None] class DateTimeSelectWidget (forms.MultiWidget): """ A widget that splits date into Date and Hours, minutes, seconds with selects """ date_format = DateInput.format def __init__(self, attrs=None, date_format=None): if date_format: self.date_format = date_format #if time_format: # self.time_format = time_format hours = [(hour,str(hour)+' h') for hour in range(0,24)] minutes = [(minute,minute) for minute in range(0,60)] seconds = minutes #not used always in 0s widgets = ( DateInput(attrs=attrs, format=self.date_format), SplitTimeWidget(attrs=attrs), ) super(DateTimeSelectWidget,self).__init__(widgets, attrs) def decompress(self, value): if value: return [value.date(), value.time()] else: [None, None, None]

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  • asp.net MVC should a View-Model Encapsulate Domain-Model?

    - by Myster
    Hi all I've see a lot of MVC examples where domain-objects are passed directly to views, this will work fine if your view is simple. The common alternative is to have a view-model which has all the same properties as your domain-model + any extra properties your view may need (such as 'confirmPassword'). Before doing too much reading and before discovering AutoMapper I started creating my own variant of view-model where the domain-object (or multiple domain objects) are simply properties of the view-model. Have I done a bad thing? What problems or benefits could be derived from this approach? Under what circumstances might this way of doing things work well?

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  • A PHP design pattern for the model part [PHP Zend Framework]

    - by Matthieu
    I have a PHP MVC application using Zend Framework. As presented in the quickstart, I use 3 layers for the model part : Model (business logic) Data mapper Table data gateway (or data access object, i.e. one class per SQL table) The model is UML designed and totally independent of the DB. My problem is : I can't have multiple instances of the same "instance/record". For example : if I get, for example, the user "Chuck Norris" with id=5, this will create a new model instance wich members will be filled by the data mapper (the data mapper query the table data gateway that query the DB). Then, if I change the name to "Duck Norras", don't save it in DB right away, and re-load the same user in another variable, I have "synchronisation" problems... (different instances for the same "record") Right now, I use the Multiton pattern : like Singleton, but multiple instances indexed by a key (wich is the user ID in our example). But this is complicating my developpement a lot, and my testings too. How to do it right ?

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  • Exposing model object using bindings in custom NSCell of NSTableView

    - by Hooligancat
    I am struggling trying to perform what I would think would be a relatively common task. I have an NSTableView that is bound to it's array via an NSArrayController. The array controller has it's content set to an NSMutableArray that contains one or more NSObject instances of a model class. What I don't know how to do is expose the model inside the NSCell subclass in a way that is bindings friendly. For the purpose of illustration, we'll say that the object model is a person consisting of a first name, last name, age and gender. Thus the model would appear something like this: @interface PersonModel : NSObject { NSString * firstName; NSString * lastName; NSString * gender; int * age; } Obviously the appropriate setters, getters init etc for the class. In my controller class I define an NSTableView, NSMutableArray and an NSArrayController: @interface ControllerClass : NSObject { IBOutlet NSTableView * myTableView; NSMutableArray * myPersonArray; IBOutlet NSArrayController * myPersonArrayController; } Using Interface Builder I can easily bind the model to the appropriate columns: myPersonArray --> myPersonArrayController --> table column binding This works fine. So I remove the extra columns, leaving one column hidden that is bound to the NSArrayController (this creates and keeps the association between each row and the NSArrayController) so that I am down to one visible column in my NSTableView and one hidden column. I create an NSCell subclass and put the appropriate drawing method to create the cell. In my awakeFromNib I establish the custom NSCell subclass: PersonModel * aCustomCell = [[[PersonModel alloc] init] autorelease]; [[myTableView tableColumnWithIdentifier:@"customCellColumn"] setDataCell:aCustomCell]; This, too, works fine from a drawing perspective. I get my custom cell showing up in the column and it repeats for every managed object in my array controller. If I add an object or remove an object from the array controller the table updates accordingly. However... I was under the impression that my PersonModel object would be available from within my NSCell subclass. But I don't know how to get to it. I don't want to set each NSCell using setters and getters because then I'm breaking the whole model concept by storing data in the NSCell instead of referencing it from the array controller. And yes I do need to have a custom NSCell, so having multiple columns is not an option. Where to from here? In addition to the Google and StackOverflow search, I've done the obligatory walk through on Apple's docs and don't seem to have found the answer. I have found a lot of references that beat around the bush but nothing involving an NSArrayController. The controller makes life very easy when binding to other elements of the model entity (such as a master/detail scenario). I have also found a lot of references (although no answers) when using Core Data, but Im not using Core Data. As per the norm, I'm very grateful for any assistance that can be offered!

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  • Making only a part of model field available in Django

    - by Hellnar
    Hello I have a such model: GENDER_CHOICES = ( ('M', 'Male'), ('F', 'Female') ) class Profile(models.Model): user = models.ForeignKey(User) gender = models.CharField(max_length=1, choices=GENDER_CHOICES) class FrontPage(models.Model): female = models.ForeignKey(User,related_name="female") male = models.ForeignKey(User,related_name="male") Once I attempt to add a new FrontPage object via the Admin page, I can select "Female" profiles for the male field of FrontPage, how can I restrict that? Thanks

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  • MVVM View Model DTOs

    - by Burt
    I have a WCF based application that uses the services to access repositories on the server side. I am passing DTOs from the server to the client and was wondering how best to make the DTOs part pf the view model. I have a workign example of just plain properties on the view model but was unsure how to deal with actual DTO objects and any possible conversion between the DTO and the Vview model properties.

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  • Django RadioButtons with icons and not label?

    - by Asinox
    Hi guy's , i have an app that have 2 fields: company_name and logo, i'm displaying the companies like radiobutton in my Form from Model, but i want to show the logo company instead of the company label (company name) Any idea ? My forms: class RecargaForm(ModelForm): compania = forms.ModelChoiceField(queryset=Compania.objects.all(), initial=0 ,empty_label='None', widget=forms.RadioSelect()) class Meta: model = Recarga Thanks :)

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  • Django FileField not saving to upload_to location

    - by Erik
    I have an Attachment model that has a FileField in a Django 1.4.1 app. This FileField has a callable upload_to parameter which, per the Django docs should be called when the form (and therefore the model) is saved. When I run FormTest below, the upload_to callable is never called and the file therefore does not appear in the location provided by the upload_to method. What am I doing wrong? Notice that in the passing tests in ModelTest (also below), the upload_to method works as expected. Test: from core.forms.attachments import AttachmentForm from django.test import TestCase import unittest from django.core.files.uploadedfile import SimpleUploadedFile from django.core.files.storage import default_storage def suite(): return unittest.TestSuite( [ unittest.TestLoader().loadTestsFromTestCase(FormTest), ] ) class FormTest(TestCase): def test_form_1(self): filename = 'filename' f = file(filename) data = {'name':'name',} file_data = {'attachment_file':SimpleUploadedFile(f.name,f.read()),} form = AttachmentForm(data=data,files=file_data) self.assertTrue(form.is_valid()) attachment = form.save() root_directory = 'attachments' upload_location = root_directory + '/' + attachment.directory + '/' + filename self.assertTrue(attachment.attachment_file) # Fails self.assertTrue(default_storage.exists(upload_location)) # Fails Attachment Model: from django.db import models from parent_mixins import Parent_Mixin import uuid from django.db.models.signals import pre_delete,pre_save from dirtyfields import DirtyFieldsMixin def upload_to(instance,filename): return 'attachments/' + instance.directory + '/' + filename def uuid_directory_name(): return uuid.uuid4().hex class Attachment(DirtyFieldsMixin,Parent_Mixin,models.Model): attachment_file = models.FileField(blank=True,null=True,upload_to=upload_to) directory = models.CharField(blank=False,default=uuid_directory_name,null=False,max_length=32) name = models.CharField(blank=False,default=None,null=False,max_length=128) class Meta: app_label = 'core' def __str__(self): return unicode(self).encode('utf-8') def __unicode__(self): return unicode(self.name) @models.permalink def get_absolute_url(self): return('core_attachments_update',(),{'pk': self.pk}) # def save(self,*args,**kwargs): # super(Attachment,self).save(*args,**kwargs) def pre_delete_callback(sender, instance, *args, **kwargs): if not isinstance(instance, Attachment): return if not instance.attachment_file: return instance.attachment_file.delete(save=False) def pre_save_callback(sender, instance, *args, **kwargs): if not isinstance(instance, Attachment): return if not instance.attachment_file: return if instance.is_dirty(): dirty_fields = instance.get_dirty_fields() if 'attachment_file' in dirty_fields: old_attachment_file = dirty_fields['attachment_file'] old_attachment_file.delete() pre_delete.connect(pre_delete_callback) pre_save.connect(pre_save_callback) Attachment Form: from ..models.attachments import Attachment from crispy_forms.helper import FormHelper from crispy_forms.layout import Div,Layout,HTML,Field,Fieldset,Button,ButtonHolder,Submit from django import forms class AttachmentFormHelper(FormHelper): form_tag=False layout = Layout( Div( Div( Field('name',css_class='span4'), Field('attachment_file',css_class='span4'), css_class='span4', ), css_class='row', ), ) class AttachmentForm(forms.ModelForm): helper = AttachmentFormHelper() class Meta: fields=('attachment_file','name') model = Attachment class AttachmentInlineFormHelper(FormHelper): form_tag=False form_style='inline' layout = Layout( Div( Div( Field('name',css_class='span4'), Field('attachment_file',css_class='span4'), Field('DELETE',css_class='span4'), css_class='span4', ), css_class='row', ), ) class AttachmentInlineForm(forms.ModelForm): helper = AttachmentInlineFormHelper() class Meta: fields=('attachment_file','name') model = Attachment UPDATE I also do testing on the Attachment model class with these unit tests -- which all pass: from core.models.attachments import Attachment from core.models.attachments import upload_to from django.test import TestCase import unittest from django.core.files.storage import default_storage from django.core.files.base import ContentFile def suite(): return unittest.TestSuite( [ unittest.TestLoader().loadTestsFromTestCase(ModelTest), ] ) class ModelTest(TestCase): def test_model_minimum_fields(self): attachment = Attachment(name='name') attachment.attachment_file.save('test.txt',ContentFile("hello world")) attachment.save() self.assertEqual(str(attachment),'name') self.assertEqual(unicode(attachment),'name') self.assertTrue(attachment.directory) # def test_model_full_fields(self): # attachment = Attachment() # attachement.save() def test_file_operations_basic(self): root_directory = 'attachments' filename = 'test.txt' attachment = Attachment(name='name') attachment.attachment_file.save(filename,ContentFile('test')) attachment.save() upload_location = root_directory + '/' + attachment.directory + '/' + filename self.assertEqual(upload_to(attachment,filename),upload_location) self.assertTrue(default_storage.exists(upload_location)) def test_file_operations_delete(self): root_directory = 'attachments' filename = 'test.txt' attachment = Attachment(name='name') attachment.attachment_file.save(filename,ContentFile('test')) attachment.save() upload_location = upload_to(attachment,filename) attachment.delete() self.assertFalse(default_storage.exists(upload_location)) def test_file_operations_change(self): root_directory = 'attachments' filename_1 = 'test_1.txt' attachment = Attachment(name='name') attachment.attachment_file.save(filename_1,ContentFile('test')) attachment.save() upload_location_1 = upload_to(attachment,filename_1) self.assertTrue(default_storage.exists(upload_location_1)) filename_2 = 'test_2.txt' attachment.attachment_file.save(filename_2,ContentFile('test')) attachment.save() upload_location_2 = upload_to(attachment,filename_2) self.assertTrue(default_storage.exists(upload_location_2)) self.assertFalse(default_storage.exists(upload_location_1))

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  • How to customize pickle for django model objects

    - by muudscope
    I need to pickle a complex object that refers to django model objects. The standard pickling process stores a denormalized object in the pickle. So if the object changes on the database between pickling and unpickling, the model is now out of date. (I know this is true with in-memory objects too, but the pickling is a convenient time to address it.) So what I'd like is a way to not pickle the full django model object. Instead just store its class and id, and re-fetch the contents from the database on load. Can I specify a custom pickle method for this class? I'm happy to write a wrapper class around the django model to handle the lazy fetching from db, if there's a way to do the pickling.

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  • Referencing Entity from external data model - Core Data

    - by Ben Reeves
    I have a external library which includes a core data model, I would like to add a new entity to this model which has a relationship with one of the entities from the library. I know I could modify the original, but is there a way to without needing to pollute the library? I tried just creating a new model with an entity named the same, but that doesn't work: * Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: 'Can't merge models with two different entities named 'Host''

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  • Alternate datasource for django model?

    - by slypete
    I'm trying to seamlessly integrate some legacy data into a django application. I would like to know if it's possible to use an alternate datasource for a django model. For example, can I contact a server to populate a list of a model? The server would not be SQL based at all. Instead it uses some proprietary tcp based protocol. Copying the data is not an option, as the legacy application will continue to be used for some time. Would a custom manager allow me to do this? This model should behave just like any other django model. It should even pluggable to the admin interface. What do you think? Thanks, Pete

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  • Model associations

    - by Kalyan M
    I have two models Library and Book. In my Library model, I have an array - book_ids. The primary key of Book model is ID. How do I create a has_many :books relation in my library model? This is a legacy database we are using with rails. Thanks.

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  • Django - Better evaluation of relationship at the model level

    - by Brant
    Here's a simple relational pair of models. class Shelf(models.Model): name = models.CharField(max_length=100) def has_books(self): if Book.objects.filter(shelf=self): return True else: return False class Book(models.Model): shelf = models.ForeignKey(Shelf) name = models.CharField(max_length=100) Is there a better (or alternative) way to write the "has_book" method? I'm not a fan of the double database hit but I want to do this at the model level.

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  • Ember model is gone when I use the renderTemplate hook

    - by Mickael Caruso
    I have a single template - editPerson.hbs <form role="form"> FirstName: {{input type="text" value=model.firstName }} <br/> LastName: {{input type="text" value=model.lastName }} </form> I want to render this template when the user wants to edit an existing person or create a new person. So, I set up routes: App.Router.map(function(){ this.route("createPerson", { path: "/person/new" }); this.route("editPerson", { path: "/person/:id}); // other routes not show for brevity }); So, I define two routes - one for create and one for edit: App.CreatePersonRoute = Ember.Route.extend({ renderTemplate: function(){ this.render("editPerson", { controller: "editPerson" }); }, model: function(){ return {firstName: "John", lastName: "Smith" }; } }); App.EditPersonRoute = Ember.Route.extend({ model: function(id){ return {firstName: "John Existing", lastName: "Smith Existing" }; } }); So, I hard-code the models. I'm concerned about the createPerson route. I'm telling it to render the editPersonTemplate and to use the editPerson controller (which I don't show because I don't think it matters - but I made one, though.) When I use renderTemplate, I lose the model John Smith, which in turn, won't display on the editTemplate on the web page. Why? I "fixed" this by creating a separate and identical (to editPerson.hbs) createPerson.hbs, and removing the renderTemplate hook in the CreatePerson. It works as expected, but I find it somewhat troubling to have a separate and identical template for the edit and create cases. I looked everywhere for how to properly do this, and I found no answers.

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  • Learning MVC - Maintaining model state

    - by GenericTypeTea
    First of all, I'm very new to MVC. Bought the books, but not got the T-Shirt yet. I've put together my first little application, but I'm looking at the way I'm maintaining my model and I don't think it looks right. My form contains the following: <% using (Html.BeginForm("Reconfigured", null, FormMethod.Post, new { id = "configurationForm" })) { %> <%= Html.DropDownList("selectedCompany", new SelectList(Model.Companies, Model.SelectedCompany), new { onchange = "$('#configurationForm').submit()" })%> <%= Html.DropDownList("selectedDepartment", new SelectList(Model.Departments, Model.SelectedDepartment), new { onchange = "$('#configurationForm').submit()" })%> <%=Html.TextArea("comment", Model.Comment) %> <%} %> My controller has the following: public ActionResult Index(string company, string department, string comment) { TestModel form = new TestModel(); form.Departments = _someRepository.GetList(); form.Companies = _someRepository.GetList(); form.Comment = comment; form.SelectedCompany = company; form.SelectedDepartment = department; return View(form); } [HttpPost] public ActionResult Reconfigured(string selectedCompany, string selectedDepartment, string comment) { return RedirectToAction("Index", new { company = selectedCompany, department = selectedDepartment, comment = comment}); } And finally, this is my route: routes.MapRoute( "Default", "{controller}/{company}/{department}", new { controller = "CompanyController", action = "Index", company="", department="" } ); Now, every time I change DropDownList value, all my values are maintained. I end up with a URL like the following after the Reconfigure action is called: http://localhost/Main/Index/Company/Sales?comment=Foo%20Bar Ideally I'd like the URL to remain as: http://localhost/Main/Index My routing object is probably wrong. This can't be the right way? It seems totally wrong to me as for each extra field I add, I have to add the property into the Index() method? I had a look at this answer where the form is passed through TempData. This is obviously an improvement, but it's not strongly typed? Is there a way to do something similar but have it strongly typed? This may be a simple-enough question, but the curse of 10 years of WinForms/WebForms makes this MVC malarky hard to get your head 'round.

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  • Logging from symfony's model layer

    - by naag
    I'm currently working on a project with symfony 1.4 and Doctrine 1.2. I'm looking for a proper way to do logging from the model layer. In some model classes I use the record hook postSave() to create a ZIP file using exec() (since PHP zip doesn't provide for storage method 'Stored'). To be sure that everythings works fine I check the return code and log an error if something goes wrong. My first naive approach was to do it like this: if ($returnCode != 0) { sfContext::getInstance()->getLogger()->debug(...); } As you know, this doesn't work so well because sfContext belongs to the controller layer and shouldn't be used from the model layer. My next try was to use the model's constructor to pass in an sfLogger instance, but this doesn't work due to Doctrine 1.2 reserving the constructor for internal use (Doctrine 1.2 Documentation). I'm looking forward for your suggestions!

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  • [CakePHP] Can not Bake table model, controller and view

    - by user198003
    I developed small CakePHP application, and now I want to add one more table (in fact, model/controller/view) into system, named notes. I had already created a table of course. But when I run command cake bake model, I do not get table Notes on the list. I can add it manually, but after that I get some errors when running cake bake controller and cake bake view. Can you give me some clue why I have those problems, and how to add that new model?

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  • WANT TO host Visual Studio Form Designer in my own application..

    - by this. __curious_geek
    Hi, I'm writing an application that lets the end use design a Form just the way Visual Studio lets you design a windows forms. I looked into visual studio forms designer articles in msdn and also studied open-source SharpDevelop editor. Both the examples host visual studio's windows forms designer to provide form-design service in their respective applications. My question is - Does it imply any licensing issue with Microsoft for directly hosting their Visual Studio Forms designer in my own application ? I don't want move ahead without proper and correct information. Can anybody here help me out if there's any licensing issues in case I want to host visual studio windows fomr designer in my own application. If there's any such licensing implication - then how does Sharp-Develop do this ?

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  • Check if Django model field choices exists

    - by Justin Lucas
    I'm attempting to check if a value exists in the choices tuple set for a model field. For example lets say I have a Model like this: class Vote(models.Model): VOTE_TYPE = ( (1, "Up"), (-1, "Down"), ) value = models.SmallIntegerField(max_length=1, choices=VOTE_TYPES) Now lets say in a view I have a variable new_value = 'Up' that I would like to use as the value field in a new Vote. How can I first check to see if the value of that variable exists in the VOTE_TYPE tuple? Thank you.

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  • Kohana PHP - Multiple apps with shared model

    - by Josamoto
    I'm using Kohana 3 to create a website that has two applications, an admin application and the actual site frontend. I have separated my folders to have the two applications separated, so the hierarchy looks as follows: /applications /admin /classes /controller /... /site /classes /controller /.... My question is, how I need to go about creating a shared /model folder. Essentially, both the admin and site itself operates on the same data, so the database layer and business logic remains more or less the same. So to me, it makes sense to have a single model folder, sitting outside of the two application folders. Is it possible to achieve the following hierarchy: /applications /model --> Where model sits in a neatly generic location, accessible to all applications /admin /classes /controller /... /site /classes /controller /.... Thanks in advance!

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  • Javascript static method intheritance

    - by Matteo Pagliazzi
    I want to create a javascript class/object that allow me to have various method: Model class Model.all() » static method Model.find() » static method Model delete() » instance method Model save() » instance method Model.create() » static that returns a new Model instance For static method I can define them using: Model.staticMethod(){ method } while for instance method is better to use: function Model(){ this.instanceMethod = function(){} } and then create a new instance or using prototype? var m = function Model(){ } m.prototype.method() = function(){ } Now let's say that I want to create a new class based on Model, how to inherit not only its prototypes but also its static methods?

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  • Creating Domain Model

    - by Zai
    Hi, I have created a use case of a small application and now I have to create a Domain Model of that use cases of the application and which functions will be implemented in this application. I have no previous experience in Domain Modeling and UML, please suggest me steps to create the domain model or any suggestions, Do I have to have a very solid understanding of Object oriented concepts for creating domain model? The application is simple and creates online poll/voting system and have functions like Register Account, Confirmation Email of account, Membership, Create Poll, Send Poll etc

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  • Why always fires OnFailure when return View() to Ajax Form ?

    - by Wahid Bitar
    I'm trying to make a log-in log-off with Ajax supported. I made some logic in my controller to sign the user in and then return simple partial containing welcome message and log-Off ActionLink my Action method looks like this : public ActionResult LogOn(LogOnModel model, string returnUrl) { if (ModelState.IsValid) { if (MembershipService.ValidateUser(model.UserName, model.Password)) { FormsService.SignIn(model.UserName, model.RememberMe); if (Request.IsAjaxRequest()) { //HERE IS THE PROBLEM :( return View("LogedInForm"); } else { if (!String.IsNullOrEmpty(returnUrl)) return Redirect(returnUrl); else return RedirectToAction("Index", "Home"); } } else { ModelState.AddModelError("", "The user name or password provided is incorrect."); if (Request.IsAjaxRequest()) { return Content("There were an error !"); } } } return View(model); } and I'm trying to return this simple partial : Welcome <b><%= Html.Encode(Model.UserName)%></b>! <%= Html.ActionLink("Log Off", "LogOff", "Account") %> and of-course the two partial are strongly-typed to LogOnModel. But if i returned View("PartialName") i always get OnFailure with status code 500. While if i returned Content("My Message") everything is going right. so please tell me why i always get this "StatusCode = 500" ??. where is the big mistake ??. By the way in my Site MasterPage i rendered partial to show long-on simple form this partial looks like this : <script type="text/javascript"> function ShowErrorMessage(ajaxContext) { var response = ajaxContext.get_response(); var statusCode = response.get_statusCode(); alert("Sorry, the request failed with status code " + statusCode); } function ShowSuccessMessage() { alert("Hey everything is OK!"); } </script> <div id="logedInDiv"> </div> <% using (Ajax.BeginForm("LogOn", "Account", new AjaxOptions { UpdateTargetId = "logedInDiv", InsertionMode = InsertionMode.Replace, OnSuccess = "ShowSuccessMessage", OnFailure = "ShowErrorMessage" })) { %> <%= Html.TextBoxFor(m => m.UserName)%> <%= Html.PasswordFor(m => m.Password)%> <%= Html.CheckBoxFor(m => m.RememberMe)%> <input type="submit" value="Log On" /> < <% } %>

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