public ActionResult DisplayView1()
EmployeeModel myModel = new EmployeeModel();
When I hit the sumit button on the view, that's method of the view is "post". How do I pass the original model back to the action?
<form id="employeeForm" name=employeeForm" action=<%Model.ActionName %> method="post">
<%Model.ActionName %> will be in this example equals "PostBackFromView"
I would like to have this:
public ActionResult PostBackFromView(EmployeeModel model)
What should I do in the view to enable the pass back of the model?
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