Overflow012 Overflow012 - 3 months ago 11
C# Question

How to show only certain columns in a DataGridView with custom objects

I have a DataGridView and I need to add custom objects to it. Consider the following code:

DataGridView grid = new DataGridView();
grid.DataSource = objects;


With this code I get a DataGridView object with all properties as columns. In my case, I don't want to show all of this information; I want to show just two or three columns. I know that I can set

AutoGenerateColumns = false
.

But I do not know how to proceed afterwards.
One option is to hide all columns that do not interest me, but I think it would be better to do it in the opposite way. How can I do this?

Answer

Whenever I do this I usually make grid.DataSource the result of a LINQ projection on the objects.

So something like this:

grid.DataSource = objects.Select(o => new
    { Column1 = o.SomeValue, Column2 = o.SomeOtherValue }).ToList();

The nice thing is that you can then set AutoGenerateColumns to true, which will generate columns based on the properties of the projected objects.

Edit:

The one downside to this approach is that by projecting everything into an anonymous object, you can have problems in situations where you need to access a specific object in a click event, for example.

In this case you may be better off defining an explicit view model and projecting your objects into those. E.g.,

class MyViewModel
{
    public int Column1 { get;set; }
    public int Column2 { get;set; }
}

grid.DataSource = objects.Select(o => new MyViewModel()
    { Column1 = o.SomeValue, Column2 = o.SomeOtherValue }).ToList();

Edit 2:

MyViewModel represents all of the columns you want to display in the DataGridView. The example properties should of course be renamed to suit what you are doing. In general, the point of a ViewModel is to serve as a sort of converter that mediates between the model (in your case your list of objects) and the view.

If you are wanting to retain a reference to the underlying object, the best way might be to supply it via the constructor:

class MyViewModel
{
    public int Column1 { get;set; }
    public int Column2 { get;set; }

    ....

    private SomeType _obj;

    public MyViewModel(SomeType obj)
    {
        _obj = obj;
    }

    public SomeType GetModel()
    {
        return _obj;
    }
}

grid.DataSource = objects.Select(o => new MyViewModel(o)
    { Column1 = o.SomeValue, Column2 = o.SomeOtherValue }).ToList();

The reason I have gone for a getter method to retrieve the underlying model object is simply to avoid a column being generated for it.