Ember save form in ajax call - ajax

I have my edit form with all of my information load in it, and when I click on the save button, it take all of field to save even if there is nothing changed in it..
I wanted to get back in an object only the index and their value that I modified in the form.
I have stricly no idea how I can do this with Ember.
Here is my code :
App.EnquiryUpdateController = Ember.ObjectController.extend({
id: null,
isSaved: false,
actions: {
save: function() {
var enquiry = this.get('model');
console.log(enquiry);
var obj = JSON.parse(JSON.stringify(enquiry));
obj = this.cleanObject(obj);
$.ajax({
url: host + 'mdf/enquiry/' + enquiry.id,
type: 'POST',
accepts: 'application/json',
data: obj
});
this.transitionToRoute('enquiry', enquiry)
}
},
});
My Route:
App.EnquiryUpdateRoute = Ember.Route.extend({
model: function() {
return this.modelFor('enquiry');
}
});
And my Template :
<script type="text/x-handlebars" data-template-name="enquiry/update">
<div class="enquiry-update">
{{#if isSaved}}
<div class="saved">Enquiry updated successfully</div>
{{/if}}
<div>
<label>Customer Name</label>
{{input value=customerName}}
</div>
<div>
<label>Customer Email</label>
{{input value=customerEmail}}
</div>
<div>
<label>Customer Phone</label>
{{input value=customerPhone}}
</div>
<div>
<button {{action "save"}}>Save</button>
</div>
</div>
</script>
If I update the customerName, I only want to have it in my object { customerName: "toto" } instead of all of them..
Thanks !

Well, the concept you're going to want to implement will be dirty attribute checking. This means you'll need to keep a copy of the original data and then either mark an attribute as dirty, or compare attributes and only send the different attributes.

Related

MVC 4.x Validate dropdown and redirect to next page

Beginner question:
I have an MVC app where there are three dropdowns on a page. Currently I'm using AJAX to evaluate a drop down on form submission and modify a CSS class to display feedback if the answer to the question is wrong.
HTML:
<form method="post" id="formQuestion">
<div class="container-fluid">
<div class="row">
<div class="col-md-4">
<p>This is a question:</p>
</div>
<div class="col-md-4">
<select id="Question1">
<option value=""></option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
</div>
<div class="col-md-4 answerResult1">
</div>
</div>
</div>
<div class="row">
<div class="col-md-6">
<button class="btn btn-success" type="submit" id="btnsubmit">Submit Answer</button>
</div>
</div>
</form>
AJAX:
#section scripts {
<script>
$(document).ready(function () {
$("#formQuestion").submit(function (e) {
e.preventDefault();
console.log($('#Question1').val())
$.ajax({
url: "/Home/DSQ1",
type: "POST",
data: { "selectedAnswer1": $('#Question1').val() },
success: function (data) { $(".answerResult1").html(data); }
});
})
});
</script>
}
Controller:
public string DSQ1(string selectedAnswer1)
{
var message = (selectedAnswer1 == "3") ? "Correct" : "Feed back";
return message;
}
I have three of these drop downs, that all get evaluated by AJAX in the same way. My question is, how would I go about evaluating all three and then returning a particular View if all three are correct.
I would like to avoid using hard-typed http:// addresses.
You could declare a global script variable prior to your document ready function, this will determine if the fields are valid. See var dropdown1Valid = false, ....
Then on your ajax success function, you could modify the values there. Say in the ajax below, your answering with first dropdown, if your controller returned Correct, set dropdown1Valid to true.
Lastly, at the end of your submit function, you could redirect check if all the variables are true, then redirect using window.location.href="URL HERE or use html helper url.action window.location.href="#Url.Action("actionName");
#section scripts {
<script>
var dropdown1Valid = false;
var dropdown2Valid = false;
var dropdown3Valid = false;
$(document).ready(function () {
$("#formQuestion").submit(function (e) {
e.preventDefault();
console.log($('#Question1').val())
$.ajax({
url: "/Home/DSQ1",
type: "POST",
data: { "selectedAnswer1": $('#Question1').val() },
success: function (data) {
$(".answerResult1").html(data);
if(data == "Correct"){
// if correct, set dropdown1 valid to true
dropdown1Valid = true;
}
// option 1, put redirect validation here
if(dropdown1Valid && dropdown2Valid && dropdown3Valid){
// if all three are valid, redirect
window.location.href="#Url.Action("actionName","controllerName", new { })";
}
}
});
// option 2, put redirect validation here
if(dropdown1Valid && dropdown2Valid && dropdown3Valid){
// if all three are valid, redirect
window.location.href="#Url.Action("actionName", "controllerName", new { })";
}
})
});
</script>
}

How to get value from radio button dynamically

i am creating a form for searching a client, using either id or email both are set to be unique. Application made on Codeignitor.
I have created a form with two radio buttons, one for search with ID and another for search with mail+dob.
Depending on the radio button selected, corresponding input fields shown.
In controller, it choose the model function based on the radio button value.
This is I coded, i need to pass the value of radio button to Controller.php file
Form(only included the radio button)
$(document).ready(function() {
$("#usingdob").hide();
$("#usingmail").hide();
$("input:radio").click(function() {
if ($(this).val() == "id") {
$("#usingId").show();
$("#usingdob").hide();
$("#usingmail").hide();
} else {
$("#usingId").hide();
$("#usingdob").show();
$("#usingmail").show();
}
});
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="col-md-4">
<label class="radio-inline">
<input type="radio" name="optradio" value="id" checked>Using ID </label></div>
<div class="col-md-4">
<label class="radio-inline">
<input type="radio" name="optradio" value="mail">Using DOB</label>
</div>
I expected to get the radio button value correctlyenter image description here
JS:
$('input[name="optradio"]').click(function(){
var optradio = $(this).val();
//or
var optradio = $("input[name='optradio']:checked").val();
if(optradio == 'id'){
//do your hide/show stuff
}else{
//do your hide/show stuff
}
});
//on search button press call this function
function passToController(){
var optradio = $("input[name='optradio']:checked").val();
$.ajax({
beforeSend: function () {
},
complete: function () {
},
type: "POST",
url: "<?php echo site_url('controller/cmethod'); ?>",
data: ({optradio : optradio}),
success: function (data) {
}
});
}
Try this
<script type="text/javascript">
$( document ).ready(function() {
$("#usingdob, #usingmail").hide();
$('input[name="radio"]').click(function() {
if($(this).val() == "id") {
$("#usingId").show();
$("#usingdob, #usingmail").hide();
} else {
$("#usingId").hide();
$("#usingdob, #usingmail").show();
}
});
});
</script>
One thing I noticed is that you have 'mail' as a value in the DOB option. Another is that there seems to be 3 options and yet you only have 2 radios?
I adjusted the mail value to dob and created dummy divs to test the code. It seems to work.
$(document).ready(function() {
$("#usingdob").hide();
$("#usingmail").hide();
$("input:radio").click(function() {
console.log($(this).val());
if ($(this).val() == "id") {
$("#usingId").show();
$("#usingdob").hide();
$("#usingmail").hide();
} else {
$("#usingId").hide();
$("#usingdob").show();
$("#usingmail").show();
}
});
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="col-md-4">
<label class="radio-inline">
<input type="radio" name="optradio" value="id" checked>Using ID </label></div>
<div class="col-md-4">
<label class="radio-inline">
<input type="radio" name="optradio" value="dob">Using DOB</label>
</div>
<div id="usingId">
Using Id div
</div>
<div id="usingdob">
Using dob div
</div>
<div id="usingmail">
Using mail div
</div>
As far as passing the value to the controller goes, ideally the inputs should be in a form. When you submit the form, the selected value can be passed to the php.
<?php
if (isset($_POST['submit'])) {
if(isset($_POST['optradio']))
{
Radio selection is :".$_POST['optradio']; // Radio selection
}
?>
If you want to get currently checked radio button value Try below line which will return current radio button value
var radioValue = $("input[name='gender']:checked").val();
if(radioValue)
{
alert("Your are a - " + radioValue);
}

AJAX Post to MVC Controller model every time empty

I am trying to send some data from a modal dialog to my controller with Ajax. But my modelfields are always null, but I enter my actionmethod in the controller.
This is a shortend version of my cshtml-file.
#model anmespace.MyModel
<form method="post" id="formID">
...
<div class="row">
<div class="col-md-5">#Resource.GetResource("MyModal", "Firstname")</div>
<div class="col-md-7"><input type="text" class="form-control" id="firstname" value="#Html.DisplayFor(model => model.FirstName)"></div>
</div>
...
<input type="submit" class="btn btn-primary" value="Submit" />
</form>
<script>
$("#formID").on("submit", function (event) {
var $this = $(this);
var frmValues = $this.serialize();
$.ajax({
cache: false,
async: true,
type: "POST",
url: "#Url.Action("ActionName", "Controller")",
data: frmValues,
success: function (data) {
alert(data);
}
});
});
</script>
Sorry MVC/Ajax are really new for me.
If you want to bind the form data to model then, the names of HTML elements should match with Model properties.
Note: name attribute value of html input field should match to the property of a model.
When you use form and submit button then it will try to reload the page by posting data to the server. You need to prevent this action. You can do this by returning false on onSubmit event in the Form element.
When you use jquery, do not forget to keep the ajax call/events inside the $(document).ready(function(){}) function.
I have written a simple code which takes First Name as input and makes an ajax call on clicking on submit button.
Html & Jquery Code:
<script>
$(document).ready(function() {
$("#formID").on("submit", function(event) {
var $this = $(this);
var frmValues = $this.serialize();
$.ajax({
cache: false,
async: true,
type: "POST",
url: "#Url.Action("PostData", "Home")",
data: frmValues,
success: function(data) {
alert(data.FirstName);
}
});
});
});
</script>
<div>
<form method="post" id="formID" onsubmit="return false;">
<input id="FirstName" name="FirstName"/>
<input type="submit" value="submit" />
</form>
</div>
My Model :
public class Person
{
public string FirstName { get; set; }
}
Action Method:
public ActionResult PostData(Person person)
{
return Json(new { Success = true, FirstName = person.FirstName });
}
Output:

generating pop up when button is clicked

when a submit button is clicked i want to generate a pop up showing the list of items. The code i tried to create pop up is as follows:`
Index View:
<script type="text/javascript">
$('#popUp').Hide();
$('#button').click(function () {
$('#popUp').click();
});
</script>
<div class="left-panel-bar">
#using (Html.BeginForm(FormMethod.Post))
{
<p>Search For: </p>
#Html.TextBox("companyName",Model);
<input id="button" type="submit" value="Submit" />
}
</div>
<div id="popUp">
#Html.ActionLink("Get Company List", "CreateDialog", "Company", null, new
{
#class = "openDialog",
data_dialog_id = "emailDialog",
data_dialog_title = "Get Company List"
});
</div>
but i got trouble using this code.. when i click the submit button it opens another page instead of popup. The controller code is as follows:
[HttpPost]
public ActionResult Index(Companies c)
{
Queries q1 = new Queries(c.companyName);
if (Request.IsAjaxRequest())
return PartialView("_CreateDialog", q1);
else
return View("CreateDialog", q1);
}
You could use AJAX:
<script type="text/javascript">
$(function() {
$('form').submit(function() {
$.ajax({
url: this.action,
type: this.method,
data: $(this).serialize(),
success: function(result) {
$('#popUp').html(result);
}
});
return false;
});
});
</script>
<div class="left-panel-bar">
#using (Html.BeginForm())
{
<p>Search For: </p>
#Html.TextBox("companyName", Model);
<input id="button" type="submit" value="Submit" />
}
</div>
<div id="popUp">
</div>
Now ehn the form is submitted, an AJAX request will be sent to the Index POST action and since inside you test if the request was an AJAX request it will return the _CreateDialog.cshtml partial view and insert it into the #popUp div. Also it is important to return false from the form submit handler in order to cancel the default even which is to redirect the browser away from the current page.

Ajax call if textarea not in form with form nesting problem as well

System I working on is CMS where you insert templates like Contact form template and save that to database. This template is coded against server side to process data.
Now my "contentDiv" within form where all the templates were insert and saved than showed on the page withint form tag wrapped like
#using (Html.BeginForm("Edit", "Home", FormMethod.Post, new { id = "first" }))
{
#Html.Hidden("someId", #Model.PageId)
}
<div id="contentDiv" style="width:100%">#Html.Raw(Model.Html)</div>
Above form is than saved as
$(function () {
$("form#first").submit(function (e) {
e.preventDefault();
var viewmodel = {
Id: $("#someId").val(),
Html: $("#contentDiv").val()
};
$.ajax({
url: $(this).attr("action"),
type: "POST",
data: JSON.stringify(viewmodel),
dataType: "json",
contentType: "application/json; charset=utf-8",
beforeSend: function () { $("#status").fadeIn(); },
complete: function () { $("#status").fadeOut(); },
success: function (data) {
var message = data.Message;
},
error: function () {
}
});
});
});
notice that I moved "contentDiv out of form tag as my contact form which is wrapped in a form tag can not be nested within id=first form.
Is there a solution to form nesting? . If not than
My another question is
contentDiv is not wrapped up in form tag that means if client browser has javascript disabled than he wont be able to post contentDiv data to server and form will be of no use.
What to do?
If I don't move contentDiv out of form tag than than after inserting template the structure will be nesting of forms
#using (Html.BeginForm("Edit", "Home", FormMethod.Post, new { id = "first" }))
{
<form id="contactform" action="/Home/Email" method="post" >
<div class="clear" style="padding-bottom:10px;"></div>
<div class="formCaption">Full Name</div>
<div class="formField"><input id="fullName" name="fullName" class="standardField" /></div>
<div><input id="sendBtn" value="Send" type="button" /></div>
</form>
}
I didn't understand from your description why the html needs to be outside the form. Also you should not use the .val() method for divs. You should use .html():
var viewmodel = {
Id: $("#someId").val(),
Html: $("#contentDiv").html()
};
Of course because you are using javascript to fetch the html which is outside of the main form if client browser has javascript disabled the form will be of no use. Only if you move the html inside the main form would this work without javascript:
#using (Html.BeginForm("Edit", "Home", FormMethod.Post, new { id = "first" }))
{
#Html.HiddenFor(x => x.PageId)
#Html.HiddenFor(x => x.Html)
<input type="submit" value="Edit" />
}
<!--
You could still keep the div for preview or something but don't make
any use of it when submitting.
-->
<div id="contentDiv" style="width:100%">
#Html.Raw(Model.Html)
</div>

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