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Angular send only changed fields of model or Replicate model with change

I prepared the form through angular.But when I need to submit the form so how can I send only all changed field or selected option values.
Need Some thing which works in all Scenario of input form fields.
DEMO: demo for ref.
on click of save how can i get it something like below one
{"list":[{"selectedOption":{"id":[]}}],"active":true,"chk_list":[{"selectedOption":{"id":["2","3"]}}],"name":"ssss"}

If you want the object to be returned to have only specific fields, then the result object should be different from the "myData" object, that you are using to actually display all the possible options to the user, etc.
Try this : http://jsfiddle.net/tc7dhep3/
I have created another variable
$scope.result
that contains the info to be sent on save button click.

The $dirty property for input elements tell if the user interacted with this, you can use this property to say which field was modified.
frm.name.$dirty
Angular doc: input - directive in module ng
You can combine this with ng-change or ng-submit to build the post object
Exemple: DEMO Forked from yours

jquery validate plugin, validating form fields of only current screen

I am using jquery validate plugin to validate form fields . I am also using jqtree . on click of every child node a section of form is visible to user, which is supposed to be filled with values.For every child there is a form content to be filled. Entire tree content is declared within one form only. I have a button in the form which on click generates json file. I am calling the function below to validate form
$("myform").validate();
....
if($("#my-form).valid())
generate the json file
but this is not validating the entire form. suppose I am on childNode1 it validates only form section defined for childNode1. As far as I have understood jquery validate plugin should validate entire form when correct form id is specified. But can anyone tell me what has gone wrong in my approach?

The .validate() method does not "validate the form". It only initializes the plugin on the form. .valid() will programmatically trigger a validation test.
Your code:
$("myform").validate();
....
if($("#my-form).valid())
generate the json file
$("myform") - Is that supposed to be an id, class, or name? As you've written it, it's looking for a <myform></myform> element.
$("#myform") // id="myform"
$(".myform") // class="myform"
$("[name='myform']") // name="myform"
Is your form element called myform or my-form? If it's the same <form> element, then the two jQuery selectors would be the same.
$("#my-form) is missing the closing quotation mark.
If the id of the <form> element is "myform", then your code should be...
$("#myform").validate(); // <- initialize the plugin
....
if ($("#myform").valid()) { // <- test the form's validity
// generate the json file
....
}
OP Title: jquery validate plugin, validating form fields of only current screen
Your question does not seem to have anything to do with the title. There is only one form described in your OP, and since this is JavaScript, only the page that's loaded in the browser is relevant. Not sure what you mean by "current screen".
but this is not validating the entire form. suppose I am on childNode1 it validates only form section defined for childNode1. As far as I have understood jquery validate plugin should validate entire form when correct form id is specified.
By default, the plugin will not validate any form fields that are hidden. You would manipulate the ignore option to over-ride this behavior. Setting ignore to [] will tell the plugin to ignore nothing and validate all fields including the hidden ones.

How to get the value of input control added through javascript

I have a button in my php view by which I can add textboxes to my page inside a form. But Whe I am getting the value of that control in my php controller the value is not there.
Is there any way by which we can get the value of dynamically added textboxes inside a POST form.

if you know the number of textbox you have, add the textbox before with a css style display:none; and display it in js on click event - for exemple with jquery
$('.yourclass').show();
it will act like a classic form
if you don't know the number you can have
you can send the form by js and get the value by something like that:
var myArray = [];
$('.yourclass').each(function() {
//unidimentionnal array exemple
myArray.push($(this).attr('myAttribut'));
});
console.log(myArray);
Seb

Collect data with PHP from dynamically created inputs

I have this code: http://jsfiddle.net/Zx8hc/9/.
It's a form where a user can add input-text dynamically, he can add and delete as many as he wants (only and the first being mandatory). And since he can add and delete the ones he likes their ID's may not be consecutive.
My question is how can I collect the data from the newly created inputs with PHP and store it in vars, each input in it's own var, independently ([input_1] > $input_1, so on). My idea was to create a for loop and go through all the elements from 0 to n, and those who contain data would be stored and those who don't wouldn't. Would that be an appropriate approach?
I am sorry that I don't have any PHP code to show but I don't even know where to start with this one, so thank you very much in advance if you can help me with this one.

I have checked your fiddle, and you have the next HTML as input element.
<input type="text">
If you want to send form data to a server, you have to wrap it in a form element. Here below is an example of a simple form
<form action="url_to_php_file.php" method="post">
<input type="text" name="age[]" />
<input type="submit" value="submit" />
</form>
Here, you see a <form> element, which you can use to pass form data to the server. It has several attributes. Here, action is the URL to where the form content should be sent to. Pick it the PHP file where the form should be handled. If it's on the same page as the display, just use # as field.
Then method-attribute is to send the form by POST data. The other option is using GET, but it's not secure because using GET will send in the form data in the URL too. While POST will wrap it in the request.
If you have a form element, it's necessary to have a button to submit the form. By submitting, you're activating the trigger to send the form to the address described in action-attribute of the form. That's the input-submit element.
Now, the data itself. Each input has to be assigned with a name-attribute. The content of it will be associated to that name when submitting the form. If you want to send in multiple data as one name field, you have to use an array, the [] in the form name.
For example, age will only hold one data-entry. While age[] can hold multiple values.
If you want to add the element, just clone the said object, only if it doesn't have id with it. If you have multiple elements with same id, you can get unpredictable results. It's advisable to keep id's unique.
And on your PHP file, read the $_POST['name'] as an array.
...... edited.

I suggest to create these new inputs with name tags. These name tags must be unique e.g. cool_input1, cool_input_2, ... OR use as array: cool_input[].
As result - you can get incoming info in php and parse received data from POST/GET.
For the first idea you don't need to know real count of the generated inputs. You just can use 'foreach element in POST' and if its name matches your pattern - this is what you need.

Checkbox creation using clone() method of jquery

I my form I have duplicate checkbox like
<input type="checkbox" name="todayDimensionStones[].isIssued" id="isIssued" value="Yes"/>
I am creating another checkbox using above using clone() method using jquery.The checkbox box is created successfully.But when I checked and submit the form containing the newly created checkbox and retrieve the value of newly created checkbox,it seems to be empty ie ''. What I to do solve this problem.If any have an idea ,please share with me

You have to change the attribute name
or you have to add [] at the end of the name.
If you send two inputs with the same name without [] at he end, php gives you the last one only.

If anyone stumbles about this: value attribute is probably missing. Set it manually after cloning:
On the cloned element with some checkboxes do:
$clone.find(':checkbox').each(function() {
$(this).attr('checked','checked')
});
Thats on IE9. See https://bugs.jquery.com/ticket/10550