Question

I am using JQuery’s UI dialog to open a form. My attempt is to submit and close the dialog. I am having trouble.

The parent window opens the dialog (which contains a partial view) from click and form is submitted, for the first time, data is saved in DB. But if I click to open the dialog again and submit, there are 2 data rows saved in DB with the same value.

My code is below:

$('#newFileDialog').dialog('destroy');$('#newFileDialog').remove();

Is there something wrong with this code?

Your help is appreciated.

UPDATE:

Partial view:

@using(Html.BeginForm("New", "File", new {}, FormMethod.Post, new { Id = "newFileForm" }  ))
{    
   @Html.ValidationSummary()    
   @Html.RenderHTML(
        new TabContainerViewModel(
                new TabViewModel { Name = "File", ViewName = "Common", Model = Model, Visible = true},
                new TabViewModel { Name = "Permission", ViewName = string.Empty, Visible = true}
            ) { Id = "createFileTabs", Visible = true })    
   @Html.SubmitButton("Send")
}
@{ Html.JQueryAjaxForm("newFileForm", "containerComponentContent", string.Empty, "$('#newFileDialog').dialog('destroy');$('#newFileDialog').remove();showMessagebox('New File added');"); }

Answer 1

I found a solution. The reason was because of live event. So, after submission, I call die event, as follow:

$('#form').die('click')

and it works!