在C#中将DataGridView的内容转换为List

时间:2010-11-25 21:54:02

标签: c# list datagridview

获取DataGridView内容并将这些值放入C#列表的最佳方法是什么?

5 个答案:

答案 0 :(得分:13)

        List<MyItem> items = new List<MyItem>();
        foreach (DataGridViewRow dr in dataGridView1.Rows)
        {
            MyItem item = new MyItem();
            foreach (DataGridViewCell dc in dr.Cells)
            { 
                ...build out MyItem....based on DataGridViewCell.OwningColumn and DataGridViewCell.Value  
            }

            items.Add(item);
        }

答案 1 :(得分:5)

如果使用DataSource绑定列表,则可以使用以下命令转换回来:

List<Class> myClass = DataGridView.DataSource as List<Class>;

答案 2 :(得分:4)

或者linq方式

var list = (from row in dataGridView1.Rows.Cast<DataGridViewRow>()
           from cell in row.Cells.Cast<DataGridViewCell>()
           select new 
           {
             //project into your new class from the row and cell vars.
           }).ToList();

答案 3 :(得分:4)

var Result = dataGridView1.Rows.OfType<DataGridViewRow>().Select(
            r => r.Cells.OfType<DataGridViewCell>().Select(c => c.Value).ToArray()).ToList();

或获取值

的字符串字典
var Result = dataGridView1.Rows.OfType<DataGridViewRow>().Select(
            r => r.Cells.OfType<DataGridViewCell>().ToDictionary(c => dataGridView1.Columns[c.OwningColumn].HeaderText, c => (c.Value ?? "").ToString()
                ).ToList();

答案 4 :(得分:1)

IEnumerable.OfType<TResult>扩展方法可以成为您最好的朋友。以下是我通过LINQ查询完成的方法:

List<MyItem> items = new List<MyItem>();
dataGridView1.Rows.OfType<DataGridViewRow>().ToList<DataGridViewRow>().ForEach(
                row =>
                {
                    foreach (DataGridViewCell cell in row.Cells)
                    {
                        //I've assumed imaginary properties ColName and ColValue in MyItem class
                        items.Add(new MyItem { ColName = cell.OwningColumn.Name, ColValue = cell.Value });
                    }
                });