I would like to get tree view using excel vba.I have many String likes this
/folderOne/fileOne
/folderTwo/fileThree
/folderOne/fileTwo
Further to the recent edit, let's say your worksheet looks like this. Note that I created some dummy samples to demonstrate duplicate sub folders.
/branches/test
/branches/test/link.txt
/branches/test/Test1/link.txt
/branches/testOne
/tags
/trunk
/trunk/test/Test1/link.txt
/trunk/testing
/trunk/testing/link.txt
/trunk/testOne
Paste the below code in a module and run it. The output will be generated in a new sheet.
CODE:
Option Explicit
Const MyDelim As String = "#Sidz#"
Sub Sample()
Dim ws As Worksheet, wsNew As Worksheet
Dim MyAr As Variant, TempAr As Variant
Dim LRow As Long, lCol As Long
Dim i As Long, j As Long, k As Long, r As Long, Level As Long
Dim delRange As Range
Dim sFormula As String, stemp1 As String, stemp2 As String
On Error GoTo Whoa
Application.ScreenUpdating = False
'~~> Set this to the relevant sheet
Set ws = ThisWorkbook.Sheets("Sheet1")
ws.Columns(1).Sort Key1:=ws.Range("A1"), _
Order1:=xlAscending, Header:=xlNo, OrderCustom:=1, _
MatchCase:=False, Orientation:=xlTopToBottom, DataOption1:=xlSortNormal
LRow = ws.Range("A" & ws.Rows.Count).End(xlUp).Row
MyAr = ws.Range("A1:A" & LRow).Value
Set wsNew = ThisWorkbook.Sheets.Add
r = 1: k = 2
With wsNew
For i = LBound(MyAr) To UBound(MyAr)
TempAr = Split(MyAr(i, 1), "/")
Level = UBound(TempAr) - 1
.Range("A" & r).Value = TempAr(1)
For j = 1 To Level
r = r + 1
.Cells(r, k).Value = Split(MyAr(i, 1), "/")(j + 1)
k = k + 1
Next j
r = r + 1
k = 2
Next
LRow = LastRow(wsNew)
lCol = LastColumn(wsNew)
For i = LRow To 1 Step -1
If Application.WorksheetFunction.CountA(.Range(.Cells(i, 2), .Cells(i, lCol))) = 0 And _
Application.WorksheetFunction.CountIf(.Columns(1), .Cells(i, 1)) > 1 Then
.Rows(i).Delete
End If
Next i
LRow = LastRow(wsNew)
For i = 2 To LRow
If .Cells(i, 1).Value = "" And .Cells(i - 1, 1).Value <> "" Then _
.Cells(i, 1).Value = .Cells(i - 1, 1).Value
Next i
For i = 2 To LRow
For j = 2 To (lCol - 1)
If .Cells(i, j).Value = "" And .Cells(i - 1, j).Value <> "" And _
.Cells(i, j - 1).Value = .Cells(i - 1, j - 1).Value Then _
.Cells(i, j).Value = .Cells(i - 1, j).Value
Next j
Next i
lCol = LastColumn(wsNew) + 1
For i = 1 To LRow
sFormula = ""
For j = 1 To (lCol - 1)
sFormula = sFormula & "," & .Cells(i, j).Address
Next j
.Cells(i, lCol).Formula = "=Concatenate(" & Mid(sFormula, 2) & ")"
Next i
.Columns(lCol).Value = .Columns(lCol).Value
For i = LRow To 2 Step -1
If Application.WorksheetFunction.CountIf(.Columns(lCol), .Cells(i, lCol)) > 1 Then
.Rows(i).Delete
End If
Next i
.Columns(lCol).Delete
lCol = LastColumn(wsNew) + 1
LRow = LastRow(wsNew)
For i = LRow To 2 Step -1
For j = lCol To 2 Step -1
If .Cells(i, j).Value <> "" And .Cells(i, j).Value = .Cells(i - 1, j).Value Then
For k = 2 To (j - 1)
stemp1 = stemp1 & MyDelim & .Cells(i, k).Value
stemp2 = stemp2 & MyDelim & .Cells(i - 1, k).Value
Next k
stemp1 = Mid(stemp1, Len(MyDelim) + 1)
stemp2 = Mid(stemp2, Len(MyDelim) + 1)
If UCase(stemp1) = UCase(stemp2) Then
.Range(.Cells(i, 1), .Cells(i, k)).ClearContents
Exit For
End If
End If
Next j
Next i
For i = LRow To 2 Step -1
If Application.WorksheetFunction.CountIf(.Columns(1), _
.Cells(i, 1).Value) > 1 Then .Cells(i, 1).ClearContents
Next i
.Cells.EntireColumn.AutoFit
End With
LetsContinue:
Application.ScreenUpdating = True
Exit Sub
Whoa:
MsgBox Err.Description
End Sub
Function LastRow(wks As Worksheet) As Long
LastRow = wks.Cells.Find(What:="*", _
After:=wks.Range("A1"), _
Lookat:=xlPart, _
LookIn:=xlFormulas, _
SearchOrder:=xlByRows, _
SearchDirection:=xlPrevious, _
MatchCase:=False).Row
End Function
Function LastColumn(wks As Worksheet) As Long
LastColumn = wks.Cells.Find(What:="*", _
After:=wks.Range("A1"), _
Lookat:=xlPart, _
LookIn:=xlFormulas, _
SearchOrder:=xlByColumns, _
SearchDirection:=xlPrevious, _
MatchCase:=False).Column
End Function
Disclaimer: I have not done any checks for /
. Please either ensure that the data has /
or put an extra line to check for /
using Instr
else you will get an error when you run the code.