C# 从 DataTable 填充 TreeView
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Populate TreeView from DataTable
提问by Azri Zakaria
I have created an account table in my database, with these columns:
我在我的数据库中创建了一个帐户表，其中包含以下列：
ID
Name
ParentID
This is an example of how records have been stored:
这是记录如何存储的示例：
ID Name ParentID
1 BANK A 0
2 0001 1
3 0002 1
4 BANK B 0
5 0001 4
6 0002 4
But the company name does not come from database, it comes from code. So how can I expand the TreeViewlike this?
但是公司名称不是来自数据库，而是来自代码。那么我怎样才能TreeView像这样扩展呢？
├─ BANK A
│ ├─ 0001
│ └─ 0002
└─ BANK B
├─ 0001
└─ 0002
How can I do this in C#? I also tried from HEREbut I still don't understand it.
我怎样才能在 C# 中做到这一点？我也从这里尝试过，但我仍然不明白。
回答by Sebastien H.
i have found a piece of code that worked great for me after spending 2h to make complicated algorithlms :
在花了 2 小时制作复杂的算法后，我发现了一段对我有用的代码：
//In Page load
//select where id is null to retrieve the parent nodes
//in a datatable (called table here)
foreach (DataRow row in table.Rows)
{
TreeNode node = new TreeNode();
node.Text = row["title"].ToString();
node.Value = row["id"].ToString();
//you can affect the node.NavigateUrl
node.PopulateOnDemand = true;
TreeView1.Nodes.Add(node);
}
then create the TreeNodePopulate event :
然后创建 TreeNodePopulate 事件：
protected void TreeView1_TreeNodePopulate(Object sender, TreeNodeEventArgs e)
{
string id= e.Node.Value;
//do your "select from yourTable where parentId =" + id;
foreach (DataRow row in table.Rows)
{
TreeNode node = new TreeNode(row["title"], row["id"])
node.PopulateOnDemand = true;
e.Node.ChildNodes.Add(node);
}
}
it worked like hell for me, i hope it will help !
它对我来说就像地狱一样，我希望它会有所帮助！
回答by Reza Aghaei
To populate a TreeViewfrom a DataTableor any IEnumerable<T>, you should be able to answer the following questions:
要从TreeViewaDataTable或 any填充 a IEnumerable<T>，您应该能够回答以下问题：
- What are data source items
- How to detect if an item in data source is a root item in tree
- How to find child items of an item in data source
- How to create tree item from data source item.
- 什么是数据源项
- 如何检测数据源中的项目是否是树中的根项目
- 如何在数据源中查找项目的子项目
- 如何从数据源项创建树项。
By passing answer of above questions as lambda expressions to the following method, it uses a recursive algorithm to create a list of TreeNodewhich you can add to TreeView. Each TreeNodecontains the descendant TreeNodeitems:
通过将上述问题的答案作为 lambda 表达式传递给以下方法，它使用递归算法创建一个列表TreeNode，您可以将其添加到TreeView. 每个都TreeNode包含后代TreeNode项目：
private IEnumerable<TreeNode> GetTreeNodes<T>(
IEnumerable<T> source,
Func<T, Boolean> isRoot,
Func<T, IEnumerable<T>, IEnumerable<T>> getChilds,
Func<T, TreeNode> getItem)
{
IEnumerable<T> roots = source.Where(x => isRoot(x));
foreach (T root in roots)
yield return ConvertEntityToTreeNode(root, source, getChilds, getItem); ;
}
private TreeNode ConvertEntityToTreeNode<T>(
T entity,
IEnumerable<T> source,
Func<T, IEnumerable<T>, IEnumerable<T>> getChilds,
Func<T, TreeNode> getItem)
{
TreeNode node = getItem(entity);
var childs = getChilds(entity, source);
foreach (T child in childs)
node.Nodes.Add(ConvertEntityToTreeNode(child, source, getChilds, getItem));
return node;
}
Example
例子
I assume you have loaded data into the following structure:
我假设您已将数据加载到以下结构中：
var dt = new DataTable();
dt.Columns.Add("Id", typeof(int));
dt.Columns.Add("Name", typeof(string));
dt.Columns.Add("ParentId", typeof(int));
dt.Rows.Add(1, "Menu 1", DBNull.Value);
dt.Rows.Add(11, "Menu 1-1", 1);
dt.Rows.Add(111, "Menu 1-1-1", 11);
dt.Rows.Add(112, "Menu 1-1-2", 11);
dt.Rows.Add(12, "Menu 1-2", 1);
dt.Rows.Add(121, "Menu 1-2-1", 12);
dt.Rows.Add(122, "Menu 1-2-2", 12);
dt.Rows.Add(123, "Menu 1-2-3", 12);
dt.Rows.Add(124, "Menu 1-2-4", 12);
dt.Rows.Add(2, "Menu 2", DBNull.Value);
dt.Rows.Add(21, "Menu 2-1", 2);
dt.Rows.Add(211, "Menu 2-1-1", 21);
Then to convert it nodes of a TreeView, you can use the following code:
然后将其转换为 a 的节点TreeView，您可以使用以下代码：
var source = dt.AsEnumerable();
var nodes = GetTreeNodes(
source,
(r) => r.Field<int?>("ParentId") == null,
(r, s) => s.Where(x => r["Id"].Equals(x["ParentId"])),
(r) => new TreeNode { Text = r.Field<string>("Name") }
);
treeView1.Nodes.AddRange(nodes.ToArray());
As a result, you will have the following tree structure:
因此，您将拥有以下树结构：
├─ Menu 1
│ ├─ Menu 1-1
│ │ ├─ Menu 1-1-1
│ │ └─ Menu 1-1-2
│ └─ Menu 1-2
│ ├─ Menu 1-2-1
│ ├─ Menu 1-2-2
│ ├─ Menu 1-2-3
│ └─ Menu 1-2-4
└─ Menu 2
└─ Menu 2-1
└─ Menu 2-1-1
