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Home/ Questions/Q 6741021
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T11:40:33+00:00 2026-05-26T11:40:33+00:00

I’ve got a datatable and one column is an integer ID foreign key to

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I’ve got a datatable and one column is an integer ID foreign key to another database table.

I’ve got a datagridview and i’d like to use a combobox column to allow user to change value. But instead of using the integers, it would be great to use the names.

I’ve tried creating a simple struct with public members int ID and string Name; a dictionary and looked into enums (however values not known at compile time) but not got anything to work yet.

I was able to populate the combobox with struct values, but not able to programmatically set the selected item/index; ie, if ID “5” is in the datatable, set the combo box selected item to the struct that has an ID of 5.

So to be clear i’m wanting:

gridview datasource's fk ID's 
1
2
3

Foreign Key table:

ID   Name 
1    Name 1
2    Name 2
3    Name 3

Datagridviewcombobox column should be loaded with three items; should display as “Name 1, Name 2, Name 3”. Based on the gridview datasource’s FK id, the selected item for each should match.

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  1. Editorial Team
    Editorial Team
    2026-05-26T11:40:34+00:00Added an answer on May 26, 2026 at 11:40 am

    You can set the DataGridViewComboBoxColumn.DataSource property, and then use the ValueMember and DisplayMember properties to determine what is shown in the ComboBox. Easiest is probably to load your FK values into DataTable and use that as the data source. For your example:

    // assuming your DataGridViewComboBox column is fkCol
    // and your fk values are in a DataTable called fkTable
    fkCol.DataSource = fkTable;
    fkCol.ValueMember = "ID";
    fkCol.DisplayMember = "Name";
    

    I am not sure how you are binding your DataGridView to your initial DataTable, but you can associate the DataGridViewComboBox column with a specific column in your original DataTable using DataPropertyName:

    fkCol.DataPropertyName = "ColumnName";
    
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