是否可以在 datagridview 的组合框列中的不同行中有不同的项目。这将使用虚拟模式。代码示例会很棒。
akantro
问问题
2574 次
1 回答
5
我想你要找的就在这里。
该技术涉及处理 DataGridView 控件的 EditingControlShowing 事件和更新 DataGridViewComboBoxEditingControl 的数据源(可能基于该行中其他列中的值)。
编辑:这是一些显示要点的代码
//Some types we'll need
enum Jobs
{
Programmer,
Salesman
}
enum DrinkCode
{
Coffee,
Coke,
MountainDew,
GinAndTonic
}
internal class Drink
{
public DrinkCode Code { get; set; }
public string Name { get; set; }
public bool Caffeinated { get; set; }
public bool Alcoholic { get; set; }
}
internal class Person
{
public string Name { get; set; }
public Jobs Job { get; set; }
public DrinkCode Drink { get; set; }
}
// the form class
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private void Form1_Load(object sender, EventArgs e)
{
BindingSource bindingSource = new BindingSource();
bindingSource.DataSource = FindPersons();
this.dataGridView1.DataSource = bindingSource;
DataGridViewComboBoxColumn column =
new DataGridViewComboBoxColumn()
{
column.DataPropertyName = "Drink";
column.HeaderText = "beverage";
column.DisplayMember = "Name";
column.ValueMember = "Code";
column.DataSource = BuildDrinksList();
}
dataGridView1.Columns.Add(column);
//handling this event is the nub of the solution
dataGridView1.EditingControlShowing +=
new DataGridViewEditingControlShowingEventHandler(dataGridView1_EditingControlShowing);
}
void dataGridView1_EditingControlShowing(object sender, DataGridViewEditingControlShowingEventArgs e)
{
//When the focus goes into the combo box cell, we can update the contents of the dropdown
//
DataGridViewComboBoxEditingControl comboBox = e.Control as DataGridViewComboBoxEditingControl;
//if you have more than one drop down this is not going to be good enough, but hey, it's an example!
if (comboBox != null)
{
BindingSource bindingSource = this.dataGridView1.DataSource as BindingSource;
Person person = bindingSource.Current as Person;
BindingList<Drink> bindingList = t his.BuildDrinksList(person);
comboBox.DataSource = bindingList;
}
}
//the rest of this is just data to make the example work
private BindingList<Drink> BuildDrinksList()
{
var drinks = new BindingList<Drink>();
drinks.Add(new Drink() { Alcoholic = false, Caffeinated = true, Code = DrinkCode.Coffee, Name = "Coffee" });
drinks.Add(new Drink() { Alcoholic = false, Caffeinated = true, Code = DrinkCode.Coke, Name = "Coke" });
drinks.Add(new Drink() { Alcoholic = false, Caffeinated = true, Code = DrinkCode.MountainDew, Name = "Mountain Dew" });
drinks.Add(new Drink() { Alcoholic = true, Caffeinated = false, Code = DrinkCode.GinAndTonic, Name = "Gin and Tonic" });
return drinks;
}
private BindingList<Drink> BuildDrinksList(Person p)
{
var drinks = new BindingList<Drink>();
if (p.Job == Jobs.Programmer)
{
drinks.Add(new Drink() { Alcoholic = false, Caffeinated = true, Code = DrinkCode.Coffee, Name = "Coffee" });
drinks.Add(new Drink() { Alcoholic = false, Caffeinated = true, Code = DrinkCode.Coke, Name = "Coke" });
drinks.Add(new Drink() { Alcoholic = false, Caffeinated = true, Code = DrinkCode.MountainDew, Name = "Mountain Dew" });
}
if (p.Job == Jobs.Salesman)
{
drinks.Add(new Drink() { Alcoholic = true, Caffeinated = false, Code = DrinkCode.GinAndTonic, Name = "Gin and Tonic" });
}
return drinks;
}
private BindingList<Person> FindPersons()
{
BindingList<Person> bindingList = new BindingList<Person>();
bindingList.Add(new Person() { Job = Jobs.Programmer, Drink = DrinkCode.Coffee, Name = "steve" });
bindingList.Add(new Person() { Job = Jobs.Salesman, Drink = DrinkCode.GinAndTonic, Name = "john" });
return bindingList;
}
}
于 2008-09-20T20:56:03.627 回答