这不仅是一个“复杂绑定”,而是一个
Pivot,您想要将重复的详细数据(cheezburgers列表)转换为一行,并且该行具有不确定数量的列。
我认为你最好的选择是编写一个自定义序列化器,使你可以将数据转换为xml数据表中的行,然后进行绑定。由于你的列数不一致,xml会更加宽容,但我不确定DataGridView会如何处理它。
编辑如下:
由于我不知道DataGridView会如何处理XML DataTable,所以我决定编写并测试它。它按照我预期的方式工作,我相信这也是你想要的。
Here are your cat & cheezburger classes (slightly modified)
public class Cat
{
public string Name { get; set; }
public string Description { get; set; }
public List<Cheezburger> Cheezbugers { get; private set; }
public void AddCheezburger(Cheezburger cheezburger)
{
if (this.Cheezbugers == null)
this.Cheezbugers = new List<Cheezburger>();
this.Cheezbugers.Add(cheezburger);
}
};
public class Cheezburger
{
public int PattyCount { get; set; }
public bool CanHaz { get; set; }
};
Then you need to create simple form with two buttons "bind to object" (button1) and "bind to datatable" (button2), with a DataGridView anchored to the bottom. and code up the form like:
//在编辑器中,下一行是在代码块中,一旦我保存它,它就不在了...
public partial class Form1 : Form
{
List<Cat> cats = new List<Cat>();
public Form1()
{
InitializeComponent();
cats.Add(new Cat() { Name = "Felix", Description = "Classic Cat" });
cats.Add(new Cat() { Name = "Garfield", Description = "Fat,Lazy" });
cats.Add(new Cat() { Name = "Tom", Description = "Wanna-Be-Mouser" });
cats[0].AddCheezburger(new Cheezburger() { CanHaz = true, PattyCount = 1 });
cats[0].AddCheezburger(new Cheezburger() { CanHaz = false, PattyCount = 3 });
cats[1].AddCheezburger(new Cheezburger() { CanHaz = false, PattyCount = 2 });
cats[1].AddCheezburger(new Cheezburger() { CanHaz = true, PattyCount = 7 });
cats[1].AddCheezburger(new Cheezburger() { CanHaz = true, PattyCount = 99 });
cats[2].AddCheezburger(new Cheezburger() { CanHaz = true, PattyCount = 5 });
cats[2].AddCheezburger(new Cheezburger() { CanHaz = false, PattyCount = 14 });
}
private void button1_Click(object sender, EventArgs e)
{
dataGridView1.DataSource = null;
dataGridView1.DataSource = cats;
}
private void button2_Click(object sender, EventArgs e)
{
dataGridView1.DataSource = null;
dataGridView1.DataSource = serializeCats(cats);
}
private DataTable serializeCats(List<Cat> cats)
{
DataTable returnTable = new DataTable("Cats");
returnTable.Columns.Add(new DataColumn("Name"));
returnTable.Columns.Add(new DataColumn("Description"));
int setID = 1;
foreach (Cat cat in cats)
{
int totalColumnsRequired = (cat.Cheezbugers.Count * 2) + 2;
while (returnTable.Columns.Count < totalColumnsRequired)
{
returnTable.Columns.Add(new DataColumn("Can Haz " + setID.ToString()));
returnTable.Columns.Add(new DataColumn("Patty Count " + setID.ToString()));
setID++;
}
returnTable.AcceptChanges();
DataRow row = returnTable.NewRow();
row[0] = cat.Name;
row[1] = cat.Description;
int cbi = 2;
foreach (Cheezburger cheezburger in cat.Cheezbugers)
{
row[cbi] = cheezburger.CanHaz;
cbi++;
row[cbi] = cheezburger.PattyCount;
cbi++;
}
returnTable.Rows.Add(row);
}
return returnTable;
}
}
不要尝试预定义DataGridView列,它们将根据数据源动态创建。绑定到猫列表会得到两列(名称/描述),绑定到数据表会获得8列,名称和描述以及6列Cheezburger信息,排列方式如您所希望的(我认为)。