Post Model back to Server - asp.net-mvc

Question

My aim is to export the clicked table, to excel (csv). This requires taking the Model on the page, and posting it to the server. But its providing difficult as its a list.
How do I post the Model in the View, to the controller.
Below is my code
View:
#model List<NameSpace.Property>
#using (Html.BeginForm("Export","Excel", FormMethod.Post))
{
<input type="submit" class="submit" value="Submit" />
}
<table .... />
Controller:
public void Export(List<Property> list)
{
}

Answer 1

As you did not state whether your table displays static data or dynamic data I have provided both solutions:
STATIC:
For example, if you request a url (lets say): /home/index/1. The table data is retrieved for that id (of 1) and then displayed in a table in the view.
If the information in the table does not change then you do not need to post any information back to the server. You should perform a GET request rather than a POST request to retrieve the CSV.
You would need to change the view to accept a model containing the original id that was requested and to use an action link rather than a form:
#model MyTableViewModel
#Html.ActionLink("Click here to export", "Export", "Excel", new { id = Model.Id });
<table .... />
Where the view model might be:
public class MyTableViewModel
{
public int Id { get; set; }
public List<Customer> Customers { get; set; }
}
public class Customer
{
public string Name { get; set; }
public string Street { get; set; }
public string PostCode { get; set; }
}
Then your export controller action would be:
public FileContentResult Export(int id)
{
string csv = //Retrieve the information for the provided id and convert to csv format
return File(new System.Text.UTF8Encoding().GetBytes(csv), "text/csv", "expostfilename.csv");
}
DYNAMIC:
The view model would no longer need the id so would become:
public class MyTableViewModel
{
public List<Customer> Customers { get; set; }
}
The view could be written:
#model MyTableViewModel
#using (Html.BeginForm("Export", "Home"))
{
<table>
<tbody>
#for (int i = 0; i < Model.Customers.Count; i++)
{
<tr>
<td>Name: #Html.TextBoxFor(m => m.Customers[i].Name)</td>
<td>Street: #Html.TextBoxFor(m => m.Customers[i].Street)</td>
<td>PostCode: #Html.TextBoxFor(m => m.Customers[i].PostCode)</td>
</tr>
}
</tbody>
</table>
<input type="submit" value="submit" />
}
And the controller export action:
public FileContentResult Export(MyTableViewModel vm)
{
string csv = //convert view model (vm) to csv
return File(new System.Text.UTF8Encoding().GetBytes(csv), "text/csv", "expostfilename.csv");
}

Answer 2

Controller Action
public void Export(List<Property> list)
{
}
Property Class
public class Property
{
public string Name { get; set; }
public string Street { get; set; }
public string Postcode { get; set; }
}
Html Table
<table id="my-table">
<tbody>
<tr>
<td>Chris</td>
<td>Awesome Street</td>
<td>DH9 4LD</td>
</tr>
<tr>
<td>Sean</td>
<td>Lame Street</td>
<td>DH8 4SR</td>
</tr>
</tbody>
</table>
Jquery
<script type="text/javascript">
var aData = {};
$('#my-table tbody tr').each(function(index, value) {
aData['list[' + index + '].Name'] = $(value).children().eq(0).text();
aData['list[' + index + '].Street'] = $(value).children().eq(1).text();
aData['list[' + index + '].Postcode'] = $(value).children().eq(2).text();
});
$.post("#Url.Action("Export")", aData, function(data) {
});
</script>
This should be enough to get you started. Hope this helps
Update
Alternatively if you did not want a Jquery specific solution, you could output your data in hidden fields
<input type="hidden" name="[0].Name" value="Chris" />
<input type="hidden" name="[0].Street" value="..." />
<input type="hidden" name="[0].Postcode" value="...." />
<input type="hidden" name="[1].Name" value="Sean" />
<input type="hidden" name="[1].Street" value="..." />
<input type="hidden" name="[1].Postcode" value="...." />
and this would get submitted in your form