Question

The reason why I am having a hard time populating this table is because the table get’s created dynamically:

MySql.Data.MySqlClient.MySqlDataReader selection = mySql.QuerySelect(textBox1.Text);  //textBox1.Text containts the text of the query to be executed

DataTable table = new DataTable();  // create table to hold results

// depending on the query construct the needed columns
for (int i = 0; i < selection.FieldCount; i++)
{
    table.Columns.Add("Column " + i);
}

// while there are rows insert them
while (selection.Read())
{
    object[] o = new object[selection.FieldCount];
    for(int j=0; j<selection.FieldCount; j++)
    {
        o[j] = selection[j].ToString();
    }
    table.Rows.Add(o);
}

ok so far after this point my table is constructed.

now my problem is how can I display that table so that user can see the results.

here are the things I have tried:

with a datagrid:

<DataGrid AutoGenerateColumns="False" Margin="73,264,17,12" Name="dataGrid1" ItemsSource="{Binding}" />

code behind:

dataGrid1.DataContext = table.DefaultView;

that does not display the results. the rows get populated and rows two but with no content…

with ListView

<ListView Height="72" Margin="78,253,17,0" Name="listView1" VerticalAlignment="Top" ItemsSource="{Binding}" />

code behind:

            listView1.DataContext = table.DataSet;
            listView1.DataContext = table.DefaultView;

note the correct number of rows are displayed but with the wrong content.

with a ListBox

similar technique….

Answer 1

You want columns in your grid, so you probably don’t want AutoGenerateColumns="False".

To use a <ListView>, you’ll need to make columns yourself in a <GridView>.

To use a <ListBox>, set DisplayTemplate to a <DataTemplate> containing UI for the listboxes.