使用 WinForms DevExpress XtraGrid c# 显示层次结构树

Question

我有分层数据结构，由以下对象组成：

Class A
{
*some properties
*list<A> SubNodes
}

现在我需要在某种可视化树中显示这个层次结构。 我正在使用 WinForms C#/ 你能告诉我如何显示这样的树吗？ Standart TreeView 没有 DataSource 属性，DevExpress XtraTreeView 使用具有 Id/ParentID 关系的 DataSource 普通源。我使用过 XtraGrid，但无法正确配置详细信息视图。你能告诉我我可以通过什么方式来处理这个问题吗？

更新

在白天的讨论中，我们决定 TreeList 不适合我们。因为我们需要提供 XtraGrid 的自动筛选行和组面板（我可以在其中拖动列进行分组）的所有东西。 因此，如果没有使用 XtraGrid 实现这种树的解决方案，我应该关闭这个问题。

Answer 1

您可以使用TreeList.IVirtualTreeListData接口。
这是一个例子：

public class A
{
    public A(string property0, string property1, string property2)
    {
        Property0 = property0;
        Property1 = property1;
        Property2 = property2;

        SubNodes = new List<A>();
    }

    public string Property0 { get; set; }
    public string Property1 { get; set; }
    public string Property2 { get; set; }

    public List<A> SubNodes { get; private set; }
}

public class TreeDataSource : List<A>, TreeList.IVirtualTreeListData
{
    void TreeList.IVirtualTreeListData.VirtualTreeGetChildNodes(VirtualTreeGetChildNodesInfo info)
    {
        var a = info.Node as A;

        info.Children = a.SubNodes;
    }

    void TreeList.IVirtualTreeListData.VirtualTreeGetCellValue(VirtualTreeGetCellValueInfo info)
    {
        var a = info.Node as A;

        switch (info.Column.FieldName)
        {
            case "Property0":
                info.CellData = a.Property0;
                break;
            case "Property1":
                info.CellData = a.Property1;
                break;
            case "Property2":
                info.CellData = a.Property2;
                break;
        }
    }

    void TreeList.IVirtualTreeListData.VirtualTreeSetCellValue(VirtualTreeSetCellValueInfo info)
    {
        var a = info.Node as A;

        switch (info.Column.FieldName)
        {
            case "Property0":
                a.Property0 = (string)info.NewCellData;
                break;
            case "Property1":
                a.Property1 = (string)info.NewCellData;
                break;
            case "Property2":
                a.Property2 = (string)info.NewCellData;
                break;
        }
    }
}

例子的用法：

var treeDataSource = new TreeDataSource();
//Add top level nodes
treeDataSource.Add(new A("Node 0, Property 0", "Node 0, Property 1", "Node 0, Property 2"));
treeDataSource.Add(new A("Node 1, Property 0", "Node 1, Property 1", "Node 1, Property 2"));
treeDataSource.Add(new A("Node 2, Property 0", "Node 2, Property 1", "Node 2, Property 2"));
//Add subnodes for first node.
treeDataSource[0].SubNodes.Add(new A("Node 0.0, Property 0", "Node 0.0, Property 1", "Node 0.0, Property 2"));
treeDataSource[0].SubNodes.Add(new A("Node 0.1, Property 0", "Node 0.1, Property 1", "Node 0.1, Property 2"));
treeDataSource[0].SubNodes.Add(new A("Node 0.2, Property 0", "Node 0.2, Property 1", "Node 0.2, Property 2"));
//Add subnodes for second node.
treeDataSource[1].SubNodes.Add(new A("Node 1.0, Property 0", "Node 1.0, Property 1", "Node 1.0, Property 2"));
treeDataSource[1].SubNodes.Add(new A("Node 1.1, Property 0", "Node 1.1, Property 1", "Node 1.1, Property 2"));
treeDataSource[1].SubNodes.Add(new A("Node 1.2, Property 0", "Node 1.2, Property 1", "Node 1.2, Property 2"));

var treeList = new TreeList();

treeList.Columns.AddField("Property0").Visible = true;
treeList.Columns.AddField("Property1").Visible = true;
treeList.Columns.AddField("Property2").Visible = true;

treeList.DataSource = treeDataSource;

Controls.Add(treeList);
treeList.Dock = DockStyle.Fill;            

示例结果：

Answer 2

有两种推荐的方法将封装树结构的业务对象绑定到 DevExpress TreeList：

1) 调整对象以在 TreeList 控件中显示。在这种情况下，对象必须实现DevExpress.XtraTreeList.IVirtualTreeListData 接口。请参阅this article，了解更多信息。
示例：How to: Implement a Tree Structure for a Business Object

2) 处理特定 TreeList 的事件并为根节点和子节点提供数据。 请参阅this article，了解更多信息。
示例：How to: Load Data Dynamically via Events

Answer 3

如果您想使用 XtraGrid 进行层次结构，则可以使用IRelationList 接口。
这是一个例子：

public class A
{
    public A(string property0, string property1, string property2)
    {
        Property0 = property0;
        Property1 = property1;
        Property2 = property2;

        SubNodes = new List<A>();
    }

    public string Property0 { get; set; }
    public string Property1 { get; set; }
    public string Property2 { get; set; }

    public List<A> SubNodes { get; private set; }
}

public class DataSource : List<A>, IRelationList
{
    IList IRelationList.GetDetailList(int index, int relationIndex)
    {
        return this[index].SubNodes;
    }

    string IRelationList.GetRelationName(int index, int relationIndex)
    {
        return null;
    }

    bool IRelationList.IsMasterRowEmpty(int index, int relationIndex)
    {
        return this[index].SubNodes.Count == 0;
    }

    int IRelationList.RelationCount
    {
        get { return 1; }
    }
}

例子的用法：

var dataSource = new DataSource();
//Add top level nodes
dataSource.Add(new A("Node 0, Property 0", "Node 0, Property 1", "Node 0, Property 2"));
dataSource.Add(new A("Node 1, Property 0", "Node 1, Property 1", "Node 1, Property 2"));
dataSource.Add(new A("Node 2, Property 0", "Node 2, Property 1", "Node 2, Property 2"));
//Add subnodes for first node.
dataSource[0].SubNodes.Add(new A("Node 0.0, Property 0", "Node 0.0, Property 1", "Node 0.0, Property 2"));
dataSource[0].SubNodes.Add(new A("Node 0.1, Property 0", "Node 0.1, Property 1", "Node 0.1, Property 2"));
dataSource[0].SubNodes.Add(new A("Node 0.2, Property 0", "Node 0.2, Property 1", "Node 0.2, Property 2"));
//Add subnodes for second node.
dataSource[1].SubNodes.Add(new A("Node 1.0, Property 0", "Node 1.0, Property 1", "Node 1.0, Property 2"));
dataSource[1].SubNodes.Add(new A("Node 1.1, Property 0", "Node 1.1, Property 1", "Node 1.1, Property 2"));
dataSource[1].SubNodes.Add(new A("Node 1.2, Property 0", "Node 1.2, Property 1", "Node 1.2, Property 2"));
//Add subnodes for second node in fisrt subnode.
dataSource[0].SubNodes[1].SubNodes.Add(new A("Node 0.1.0, Property 0", "Node 0.1.0, Property 1", "Node 0.1.0, Property 2"));
dataSource[0].SubNodes[1].SubNodes.Add(new A("Node 0.1.1, Property 0", "Node 0.1.1, Property 1", "Node 0.1.1, Property 2"));
dataSource[0].SubNodes[1].SubNodes.Add(new A("Node 0.1.2, Property 0", "Node 0.1.2, Property 1", "Node 0.1.2, Property 2"));

var gridControl = new GridControl();

var view = new GridView(gridControl);           

gridControl.MainView = view;
gridControl.DataSource = dataSource;
gridControl.Dock = DockStyle.Fill;

view.OptionsDetail.ShowDetailTabs = false;

Controls.Add(gridControl);

示例结果：

另外，您可以查看«How to emulate a TreeList using the master-detail GridView» 示例。