五级目录树的建立 - 专家坐堂 - Foxtable(狐表) - 新一代数据库软件，完美融合Access、Foxpro、Excel、vb.net之优势，人人都能掌握的快速软件开发工具!

“编辑目录树之二”是三级目录树的建立，请问五级目录树的下列代码应如何修改

5、“增加子节点”按钮的Click事件代码：

Dim tr As WinForm.TreeView
Dim nd As WinForm.TreeNode
tr = e.Form.Controls("TreeView1")
nd = tr.SelectedNode
If nd IsNot Nothing Then
    If nd.Level = 2 Then
        MessageBox.Show("最多允许三层节点!","提示",MessageBoxButtons.OK,MessageBoxIcon.Information)
    Else
        Dim dr As DataRow = DataTables("表A").AddNew
        If nd.Level = 0 Then
            dr("大类") = nd.Text
            dr("二类") = "新节点"
        ElseIf nd.Level = 1
            dr("大类") = nd.ParentNode.Text
            dr("二类") = nd.text
            dr("三类") = "新节点"
        End If
        nd = nd.Nodes.Add("新节点")
        tr.SelectedNode = nd
        tr.Select()
        tr.BeginEdit()
    End If
End If

6、“删除”按钮的Click事件代码：

Dim tr As WinForm.TreeView = e.Form.Controls("TreeView1")
Dim nd As WinForm.TreeNode = tr.SelectedNode
If nd IsNot Nothing Then
    Dim pth() As String = nd.FullPath.Split("\")
    Select Case nd.Level
    Case 0
        DataTables("表A").DeleteFor("大类 = '" & pth(0) & "'")
    Case 1
        DataTables("表A").DeleteFor("大类 = '" & pth(0) & "' And 二类 = '" & pth(1) & "'")
    Case 2
        DataTables("表A").DeleteFor("大类 = '" & pth(0) & "' And 二类 = '" & pth(1) & "' And 三类 = '" & pth(2) & "'")
    End Select
    nd.Delete()
End If
tr.Select()

7、目录树AfterEditNode事件代码：

If e.NewText = "" Then
    e.Cancel = True
    Return
End If
Dim pth() As String = e.Node.FullPath.Split("\")
Select Case e.node.Level
    Case 0
        DataTables("表A").ReplaceFor("大类",e.NewText,"大类 = '" & pth(0) & "'")
    Case 1
        DataTables("表A").ReplaceFor("二类",e.NewText,"大类 = '" & pth(0) & "' And 二类 = '" & pth(1) & "'")
    Case 2
        DataTables("表A").ReplaceFor("三类",e.NewText,"大类 = '" & pth(0) & "' And 二类 = '" & pth(1) & "' And 三类 = '" & pth(2) & "'")
End Select
e.Node.Name = e.NewText

老师我下面的代码如何修改，仅生成一条数据。而不是生成多条数据再删除

Dim dt As DataTable = DataTables("department")

Dim dic As New Dictionary(of Integer, Integer)

Dim prevLevel As Integer = 0

Dim idxs As String = "-1,"

For Each nd As WinForm.TreeNode In e.Form.Controls("TreeView1").AllNodes

If nd.Level = 0 AndAlso dic.count > 0 Then

Dim temp = dic(0)

dic.Clear

dic.Add(0, temp)

End If

If dic.ContainsKey(nd.Level) = False Then

dic.Add(nd.Level, 1)

Else

dic(nd.Level) = dic(nd.Level) + 1

End If

For i As Integer = nd.level + 1 To prevLevel

dic(i) = 0

prevLevel = nd.Level

Dim ndr As DataRow

If nd.DataRow IsNot Nothing Then

ndr = dt.Find("_Identify = " & nd.DataRow("_Identify"))

Else

ndr = dt.AddNew

End If

idxs &= ndr("_Identify") & ","

Dim str As String = ""

For i As Integer = 0 To nd.Level

str &= format(dic(i), "000") & "."

ndr("departno") = str.TrimEnd(".")

ndr("department") = nd.Text

' dt.DeleteFor("_Identify not in (" & idxs & ")")

蓝老师，这种方法目录树的层级就收到限制了。我这边是想做部门的目录层级。部门目录树估计能达9级左右。

[此贴子已经被作者于2020/5/30 8:53:38编辑过]

我现在的想的最优方案，一条数据包含上级部门这个字段。通过上级部门这个字段能否生产无限极目录树？

老师生产目录树的表和编辑下拉窗口所在的表是同一张表。

在NodeMouseClick设置下列代码，点击目录树后提示“未将对象引用设置到对象的实例。”

Dim tr As Row = Tables("department").Current

Dim dr As DataRow = e.Node.DataRow '获取生成此节点的DataRow

tr("parent") = dr("departname")

tr("parentID") = dr("departID")

e.Form.DropDownBox.Value = tr("departname") '这行不是多余的,省略此行,无法在表中正常使用此下拉窗口输入数据.

e.Form.DropDownBox.CloseDropdown()

老师修改后点击某个目录树某个节点没反应。Dim tr As Row = Tables("department").Current ，department表取数据生产的目录树tree1。现在点击department的上级部门列，出来下拉窗口中的目录树就是tree1。

这样是不是不行？

就是希望点击部门信息表中的“上级部门”可以达到代码中的赋值

老师还希望目录树中的排列顺序按排序号值进行排序。

事列：

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点击浏览该文件:测试.zip

[此贴子已经被作者于2020/5/30 15:48:54编辑过]

好的，谢谢

[此贴子已经被作者于2020/5/30 16:23:06编辑过]

主题：五级目录树的建立