D
Daniel
Hello,
Within VB6, I have Excel(2000) open and I need to open another workbook
using Application.GetOpenFilename(). I am using ChDir and ChDrive before.
Eventhough ?App.Path in the immediate windows shows the correct path, when
executing the method, another drive and folder is showned in the dialogue
box. Is there a way to get it to open where I would like it to open?
Your help will be appreciated.
Sub prepareTSData()
Dim selectedFile As String, saveInThisDir As String
Dim theDrive As String, theFolder As String
'display dialog asking user to select a TSData file
saveInThisDir = saveInDir(currentFileInfo.fileName)
theDrive = Left(fullPath, 1)
ChDrive theDrive
theFolder = currentFileInfo.fileDir
ChDir theFolder
selectedFile = Application.GetOpenFilename( _
"Excel Files (*.xls),*.xls", , _
"Select your TSData file", , False)
............
Daniel
Within VB6, I have Excel(2000) open and I need to open another workbook
using Application.GetOpenFilename(). I am using ChDir and ChDrive before.
Eventhough ?App.Path in the immediate windows shows the correct path, when
executing the method, another drive and folder is showned in the dialogue
box. Is there a way to get it to open where I would like it to open?
Your help will be appreciated.
Sub prepareTSData()
Dim selectedFile As String, saveInThisDir As String
Dim theDrive As String, theFolder As String
'display dialog asking user to select a TSData file
saveInThisDir = saveInDir(currentFileInfo.fileName)
theDrive = Left(fullPath, 1)
ChDrive theDrive
theFolder = currentFileInfo.fileDir
ChDir theFolder
selectedFile = Application.GetOpenFilename( _
"Excel Files (*.xls),*.xls", , _
"Select your TSData file", , False)
............
Daniel