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How to send model data from one controller to another?
Should I use session or should I save it to db? You should keep these as last options. Following are my suggestions. TempData RedirectToAction TempData - This is just like a DataReader. Once the key is read the data will be become null. In case you still want to persist the data after reading it you can use the code like below TempData.Alive() Alive function will persist the data even after the TempData for the particular key is read. RedirectToAction - In this approach you can set the Query String Values in the parameter list. Below is the sample code. return RedirectToAction( "Action Name", new RouteValueDictionary ( new {

Categories : Asp Net Mvc

How to send back the model data from jsp to controller
You need to make them into form inputs using the Spring form tags in much the same way as you have for the form:select. If they are not editable by the user, you can always disable them.

Categories : Spring

How to pass Model and some form data to controller?
There are so many things wrong with your code that I don't know where to begin with. You don't post your Model back to the Controller using an ActionLink. You should use a submit button in your form to submit it to the server (Controller). Use Html Helpers instead of HTML tags. Your ViewModel should be under ViewModels folder and namespace, not under Controllers. And, if you want to submit additional data along with your Model, you should add them as properties to your View Model. Here is what your View should look like: @model BitcoinRedeemPage.ViewModels.BitcoinTransactionViewModel @using(Html.BeginForm("Send", "DepositDetails")) { @Html.TextBoxFor(model => model.WallterAddress) <input type="submit" name="submit" value="Submit" /> } And, in your Controller you'l

Categories : Asp Net Mvc

MVC4 passing all form data (model) from javascript to controller -- is it possible?
Basically, you can do it calling an AJAX POST: JS (using jQuery): $('form').on('submit', function (event) { // Stop the default submission event.preventDefault(); // Serialize (JSON) the form's contents and post it to your action $.post('YourAction', $(this).serialize(), function (data) { // If you want to do something after the post }); }); Controller Action: public ActionResult YourAction(string JSONmodel) { System.Web.Script.Serialization.JavaScriptSerializer serializer = new System.Web.Script.Serialization.JavaScriptSerializer(); MyModel model = serializer.Deserialize(JSONmodel, typeof(MyModel)); // Now you can do whatever you want with your model } UPDATE For more complex objects, you can use a third part solution for serialization/de

Categories : Javascript

Sails.js - issues when trying to display model data
Likely, your browser is stuck because an error happens when you're trying to find an Applicant, and your code doesn't return any response in this case. So browser waits for response forever. Please try something like this if (err) { console.log('HAI'); return res.send(err, 500); } P.S. By the way, as of Sails v0.9, find() method will alwals return an array, even if only one record is found. If you want to find just one record by id, and expect single object in your view, you can use findOne() method.

Categories : Javascript

Sails.js: How To Access Data in Server-Side EJS View for Single Model
You can create controller action that matches the route and manually pass found model to the view. In UserController: find: function(req, res) { User.findOne({'id': req.params['id']}, function(err, user) { res.view({user: user}) }) } Then You can reference this model in views/user/find.ejs: <%- user.id %> <%- user.name %>

Categories : Misc

Should i send a mail from a model callback or from a controller?
Why would it be bad? Use ActiveRecord::Observer for that. It's a perfect use case, because mail logic probably doesn't belong neither to your model, nor controller. class WelcomeEmailObserver < ActiveRecord::Observer observe :user def after_create(user) if user.purchased_membership? GreetingMailer.welcome_and_thanks_email(user).deliver else GreetingMailer.welcome_email(user).deliver end end end # app/models/user.rb class User end # config/application.rb class Application < Rails::Application config.active_record.observers = :welcome_email_observer end Example stolen from here. I would also suggest that you leave your observers stateless, because it's hard to debug them otherwise. Use them only for callbacks with third-party API's, emails or s

Categories : Ruby On Rails

How to send a model property value from a view to controller
This line: @Html.HiddenFor(id => id.ItemId, new { @class = "id" }) needs to be inside your Using statement. The same goes for your submit button/input. In order for the ajax form to submit properly those two must be inside the Using block.

Categories : Asp Net Mvc

Sails.js server not starting anymore
Hmm-- looks like port 8080 isn't available. What happens if you try to switch the port? You may have another server running on that port. Or in some cases, hosts require the hostname to be set. I'd try switching the port first though.

Categories : Javascript

Send a model to MVC3 controller by setting src of element
Try to use an html form with it's target attribute set to the name of your iframe, something like: <iframe style="display: none;" id="myIframe" name="myIframe"></iframe> var f = document.createElement("form"); f.method = "POST"; f.action = "../Workflow/ExportPatchSchedules"; f.target = "myIframe"; f.enctype = "application/x-www-form-urlencoded"; // not sure about this since you didn't mention var input = document.createElement('input'); input.type = "hidden"; input.name = "model"; input.value = "...." // your data here, not sure about it since you didn't mention f.appendChild(input); f.submit();

Categories : C#

How can send List<> Model From View To Controller using Ajax Jquery in Mvc3 asp.net?
It's a Model Bind List, in this case, you have to send the information using the same name of your parameter in the name attribute of html inputs tag and asp.net model binder will convert it on a collection on the post. Your javascript looks fine. Try something like this: In the view: <input type="text" name="roles" value="1" /> <input type="text" name="roles" value="4" /> <input type="text" name="roles" value="2" /> <input type="text" name="roles" value="8" /> in the controller: public ActionResult Post(List<int> roles) { // process } Also take a look at this article: http://haacked.com/archive/2008/10/23/model-binding-to-a-list.aspx

Categories : Asp Net Mvc

PHP: Send form data to php script which then posts data to another form processor like zapier.com?
Using javascript and Ajax it can be done like this (with jQuery): $("#idOfTheForm").submit(function(e) { e.preventDefault(); $.ajax({ method : "post", url : this.action, data : $(this).serialize(), success : function() { window.location = "yourUrlOfThanks.html"; }, error : function() { alert("Something went bad"); } }); }); So basically it is: sent a post request to the action url of this form, and once it throws an 200 code (found and everything went ok) then redirect to my page. If soemthing went wrong, then the server will throw an 40* status code and the script will go into error function.

Categories : PHP

Updating repository pattern model by sending ViewModel to controller - How to send back to context?
you can read my code, i use repository pattern to.. this my controller [Authorize] public ActionResult EditUser(int UserID) { List<UserViewModel> UserViewModel = _AccountService.userViewModel(UserID); return View(UserViewModel.FirstOrDefault()); } [Authorize] [HttpPost] public ActionResult EditUser(UserViewModel updateUser) { var UpdateData = _AccountService.UpdateDataUser(updateUser.UserID, updateUser.FullName, updateUser.Email, updateUser.IsActive, updateUser.IsMaster); if (UpdateData != null) { return RedirectToAction("Users"); } return View(UpdateData); } this my service class public List<UserViewModel> us

Categories : Asp Net Mvc

Send parameters to controller from view with no form/input
Do an AJAX request when the button is clicked. $.ajax({ type: "POST", url: http://site.com/Controller/Action, data: data, success: success, dataType: dataType }); See: http://api.jquery.com/jQuery.post/

Categories : C#

Send back value from controller when submit form codeigniter
validate the data that comes from your sign in form. Never pass data from a form directly into a database query. validating the data will tell you if the form has correct values. if it does not, then show the form again. form data is validated - query your database there are 3 possible results - you got a match OR you did not get a match OR there was an error. Check for these 3 different conditions, and then show the appropriate view.

Categories : Codeigniter

How to send form values to the route of the controller using Symfony2
You can submit form and get form data in controller(not need to specify it in url) {% block body %} <form action="{{ path('advd_status') }}" method="post" {{ form_enctype(form) }}> {{ form_widget(form) }} <input class="input_button" type="submit" value="Search" /> </form> {% endblock %} In controller public function groupPrizesRequestedAction($status = 0, $country = 0) { $formulario = $this->createForm(new GroupingType()); if ($request->getMethod() == 'POST') { $data = $formulario->getData(); $status = $data->getStatus(); switch ($status) { ... } } else { return $this->render('ADVDBundle:Prize:groupIndex.html.twig', array('form' => $formulario->createView());

Categories : Forms

data from model to controller and back to model
get_conditioncurrently returns an array; try the code below: function get_condition() { $user = $this->session->userdata('user_id'); $this->db->select('condition'); $this->db->from('staff'); $this->db->where('user_id',$user ); $query = $this->db->get(); $result = $query->row_array(); return $result['condition']; } Edit: looks like your $user var in that where clause is also an array.

Categories : PHP

SelectBox with form controller and remote model
Virtual select box does not help you much because it's only virtual when it comes to rendering. So the data binding and the race conditions should be the same no matter which select box you choose. I don't get the root of your problem so I can't give you a precise advice how to handle your race condition. Is it possible to load / set the models sequentially to break the race condition?

Categories : Misc

Spring form send null model object to database
Try adding @modelattribute in this method .It fill fetch the required model object. @RequestMapping(method = RequestMethod.POST) public String addSpitterFromForm(**@ModelAttribute("spitter")** @Valid Spitter spitter, BindingResult bindingResult) { if(bindingResult.hasErrors()) return "spitters/edit"; spitterService.saveSpitter(spitter); return "redirect:/spitters/" + spitter.getUsername(); } and just to check if it is getting the values from form,syso some values like syso(spitter.getUserName) to check if values are coming. ALso, I believe that you are making a constructor and passing service to it ,so there is no need of @Inject @Inject //@Autowired///Why are you injecting it if it is a constructor? public SpitterController(SpitterService spitterService) {

Categories : Java

AngularJS ng-model form driven by ng-repeat over UI model description data how to?
I've just figured out one (but may be not the best?) way to achieve this myself.. see http://jsfiddle.net/vorburger/8CxRC/3/ - basically still based on my square bracket dynamic keys 'trick', but with some pre-processing: for (var i = 0; i < $scope.uimodel.length; i++) { var resolvedModelProperty = $scope.datamodel; var remainingPath = $scope.uimodel[i].model; while (remainingPath.indexOf('.') > -1) { var nextPathSlice = remainingPath.substr(0, remainingPath.indexOf('.')); resolvedModelProperty = resolvedModelProperty[nextPathSlice]; remainingPath = remainingPath.substr(remainingPath.indexOf('.') + 1); } $scope.uimodel[i].modelB = resolvedModelProperty; $scope.uimodel[i].modelL = remainingPath;

Categories : Javascript

Trying to send input file data over AJAX, can't access the data in my controller
Going off of what Kevin B commented, there are a couple ways to do this. First, the reason it is not working is that, by default, you cannot send files with an AJAX request. That is it, and that is why it isn't working. No matter what you do to your form and your AJAX request, you are stuck. (AJAX here meaning NOT XMLHttpRequest2) SOLUTION 1 Kevin B recommended the Javascript formData object which is part of the XMLHttpRequest Level 2. Information on how to use it can be found: https://developer.mozilla.org/en-US/docs/Web/Guide/Using_FormData_Objects In relation to your code, you code do something along the lines of: commentforms.on('submit', function(e){ e.preventDefault(); var oData = new FormData(document.forms.namedItem("composeForm")); var oReq = new XMLHttpRequest(

Categories : Javascript

How to use `calloutAccessoryControlTapped` is touched correctly to send data from one view controller to a detail view controller
detailViewController.listingId = theAnnotation.catListingMapId and in detailViewController find the other details by listingID EDIT: in @interface MyAnnotation write a method: - (NSDictionary *) getTheAnnotationData{ NSDictionary* theDict = [NSDictionary dictionaryWithObjectsAndKeys: [NSString stringWithFormat:@"%f",latString],@"latString", [NSString stringWithFormat:@"%f",lngString],@"lngString", [NSString stringWithFormat:@"%i",catMapId],@"catMapId", [NSString stringWithFormat:@"%i",catListingMapId],@"catListingMapId", title,@"title", subtitle,@"subtitle", nil]; return t

Categories : Iphone

How do I direct the routing of a form to a controller other than the one associated with the model class?
If your view has multiple models and you want to update particular model then you can pass that particular model or may be just id and changed attributes through url in params hash. But it mostly depends on what exactly you are trying to achieve. May be you should rewrite your exact problem to make it clear.

Categories : Ruby On Rails

Laravel: Form model binding and resource controller error
You have a problem with the form action. Assuming you have a route like this: Route::post('task/update/{id}, function() { }); Then, your model-bound form should be: {{ Form::model($task, array('url'=>array('task/update',$task->id),'files'=>true)) }}

Categories : PHP

Passing Model in ViewModel to controller get null (Edit Form)
You have: @Html.DropDownList("BranchID") You want: @Html.DropDownListFor(model => model.caseInternal.BranchID) OR if you really don't want to use a DropDownListFor() for whatever reason.... @Html.DropDownList("caseInternal_BranchID")

Categories : Asp Net Mvc

Submitting Form in MVC 4 model is null in controller post method
Your View is type of BillingCodeList, but you are trying to send List<BillingCodeObj> to you action in controller. Try to change action as follows: [HttpPost] public void SubmitTimesheet(BillingCodeList model) { string uri = ""; foreach (var billingCode in model.BillingCodes) { //do stuff with each of these } }

Categories : C#

Controller computed property is not showing new data additions immediately, where the data is not those returned from the route model
The problem is that the computed property is cached and will only refresh when office.@each changes. The office property is not defined on that controller, so office.@each is always null. Probably what you want instead is content.@each. So: officelist: function(){ return App.Office.find({org_id: 'myorgid'}); }.property('content.@each') Now the page will refresh whenever a new office is added.

Categories : Ember Js

How to use ModelState to clear my form after I post my form data,my view model is a viewmodel which has two models?
In such case you will have to remove each property of StudentModel from your ModelState i.e. Syntax: ModelState.Remove("PropertyName"); Example: ModelState.Remove("Id"); ModelState.Remove("Name"); ModelState.Remove("Marks"); Edit: For specific model property in among two ModelState.Remove("StudentModel.Name");

Categories : Asp Net Mvc

Laravel 4: Can't get model data with a form macro on edit form
For reference if you want to check what the Form generates you can look up vendor/laravel/framework/src/Illuminate/Html/FormBuilder. In the FormBuilder class you'll find methods for the inputs. The method for the text input looks like this public function text($name, $value = null, $options = array()) { return $this->input('text', $name, $value, $options); } As you can see the second parameter is the value and the third is the options array. So for your form it would look like this Form::text('email', null, array('class'=>'m-wrap span12'))

Categories : PHP

Rails. Form submit contains nested model, call method from models controller
Can you try this? def create @foo = Foo.new(params[:foo]) @foo.bars[0] = request.remote_ip if @foo save #it's ok end end Explanation: when you have nested attributes, you create Foo and one or more Bars (one on many relationship). In your case, you make only one Bar, but that doesn't change the fact you have one on many relationship, you just have one element in bars. So, you put bars[0] to access first and only element of your bars array (ActiveRecord::Relation).

Categories : Ruby On Rails

sails.js: how to update model
So you figured out your problem, sort of. req.body is already an object. But you really should sanitize it before you put it into your update and then save the object. There's a lot of reasons for this but with Mongo when you get only a partial object you'll replace the object in the collection which, in your example with a user, could be bad. When I send users to the frontend I cull off things I don't want transmitted all over like passwords. The other reason is the golden rule of web application development - never trust the client! I'd start with something like: var user = User.findOne(req.body.id).done(function(error, user) { if(error) { // do something with the error. } if(req.body.email) { // validate whether the email address is valid? // Then

Categories : Json

Data from model to controller
Lifetime - DbContext The lifetime of the context begins when the instance is created and ends when the instance is either disposed or garbage-collected. Use using if you want all the resources that the context controls to be disposed at the end of the block. When you use using, the compiler automatically creates a try/finally block and calls dispose in the finally block. The problem was when the inner using got disposed, it invalidated the DbContext. So you need to use .ToList() to save the query result in memory. Suppose getChatLogWithName is defined in the class called Repo, you can change the controller logic to something like this: var repo = new Repo(); var items = repo.getChatLogWithName().ToList(); Or move .ToList() to getChatLogWithName. Btw, you should not

Categories : C#

How to send data from view to controller on button click
In order for the data to be submitted to the controller, the inputs must appear within the <form> tag. For example: <body> <div> @using (Html.BeginForm("FindPerson", "MyController", FormMethod.Post)) { @Html.DropDownList("Name") <br /> @Html.DropDownList("Age") <br /> @Html.DropDownList("Gender") <br /> <input type="submit" value="Find" /> } </div> </body>

Categories : C#

Jquery doesn't send data to Spring controller
@PathVariable is for information embedded in the URL 'REST Style'. e.g.,: @RequestMapping(value="/test/user/{login}/password/{password}/email/{email}", method=RequestMethod.POST) To read in regular old http parameters that are in the query string or post body the annotaton is @RequestParam, rather than @PathVariable. public void test(@RequestParam("login") String login, @RequestParam("password") String password, @RequestParam("email") String email) { Also in your ajax, data is expecting, well, data. If you want to format it as query string, it should be appended to the URL directly. to put it in the body you would do: data: {login:login ,email: email,password: password} rather than format it as a query string. You can also just use jQuery's .serialize() method on the form tag

Categories : Java

Laravel 4 form data submit and update the database and also send an email with the help of post data
For the most part everything looks good except in your SendmailController, method post_create you have an error. The ";" should be outside of the last ")" $new_mail = Sendmail::create( array( 'recipients' => Input::get('toaddress'), 'subject' => Input::get('subject'), 'body' => Input::get('body') ) ); Are you getting the error on the posting of the form or just showing the form?

Categories : PHP

Calling super method in sails.js controller
Update: for Sails >= v0.10.x, see the comment below from @naor-biton If you want to access the default implementation (the blueprint), as of v0.9.3, you can call next() (the third argument to your controller). This is because Sails is based on the Express/Connect concept of middleware, allowing you to chain things together. Please note that this behavior may change in a subsequent version, since next() is also how you call your default 404 handler (config/404.js) for actions which don't have a blueprint underneath them. A better approach, if you're interested in using the blueprints but running a bit of logic beforehand, is to leave the controller action undefined and use one or more policies, which will be run beforehand.

Categories : Javascript

How to make can.Model send JSON data when update?
The update method signature for using a function is can.Model.update: function(id, serialized) -> can.Deffered. So as the first parameter you always get the id and as the second the serialized data. Changing your update : function(story) {} to update : function(id, story) {} should solve the problem.

Categories : Javascript

Send back data from activerecord model on create
Only way I can see is to: after_initialize you assing key to additional attr_accessor (let's call it unhashed_key) add before_validation on: :create that will take unhashed_key, hash it and assign to hashed_key On save only hashed_key goes to db and unhashed_key is still available for reading You pass unhashed_key through flash or session for next request (I don't think flash is the best way, use session maybe)

Categories : Ruby On Rails

Sails.js - Creating a view to edit a model
You should point browser to localhost:1337/human/edit/:id url, where :id is your particular model's id. For example <ul> <% _.each(humans, function(model) { %> <li><%= model.firstName %> <a href="/human/edit/<%= model.id %>"><button id="<%=model.firstName %>" type="button">Edit</button> </a> </li> <% }) %> </ul> This will automatically execute Human.edit controller with id param set to particular model's id. You don't have to write any custom routes, this is default behaviour. Example of Human/edit controller action: edit: function(req, res) { Human.findById( req.param('id') ) .done(function(err, human) { if (err) { return res.send(err, 500); } else { if (human.length)

Categories : Javascript

Why unsetting data in model does not affect controller?
This is basic OOP. If you do not pass objects along (but data arrays), you cannot - by itself - expect pass by reference. Therefore modification to the data in your model cannot also alter the data in your request object. After you passed them they are independent. If you need - for some reason - to update your request object, you need to pull the data again: if ($this->Model->save($this->request->data)) { // redirect on success? } $this->request->data = $this->Model->data;

Categories : PHP



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