$.post("Yourcontroller/YourAction", { FirstName : $("#txtFirstName").val(), LastName : $("#txtLastName") } ,function(data){
//do whatever with the response
});
Your ViewModel Property names and Parameter we are passing should be same. Ie : Your view model should have 2 properties called FirstName and LastName like his
public class PersonViewModel
{
public string FirstName { set;get;}
public string LastName { set;get;}
// other properties
}
And your Post action method should accept a parameter of type PersonViewModel
[HttpPost]
public ActionResult YourAction(PersonViewModel model)
{
//Now check model.FirstName
}
Alternatively, If your view is strongly typed to the PersonViewModel, you can simply send the serialized form to the action method using jQuery serialize method
$.post("Yourcontroller/YourAction", $("#formId").serialize() ,function(data){
//do whatever with the response
});
EDIT : As per the comment
Serialize will take care of the Child property as well. Assume you have a class called Profession like this
public class Profession
{
public string ProfessionName { set; get; }
}
And your PersonViewModel has a property of type Profession
public class PersonViewModel
{
//other properties
public Profession Profession { set; get; }
public PersonViewModel()
{
if (Profession == null)
Profession = new Profession();
}
}
You will get these data in your HttpPost Action method, if you fill that from your view.
