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Posted

heloo everyone

i am facing error locating a file

i am having two forms

i am accepting a user's photograph in form one.

after form one is saved and i open form two where i am access a text file

which is located in the directory where the software exe is running

but it searches the text file in direcoty where the jpeg was selected from.

 

i think there is some initialisation problem.

can any one help me

 

or is there any command where i can tell it to read the file from current directory itself

 

the cade that i am using is

 

'form1

Private Sub OpenPicDialog()

Dim OpenDialog As New OpenFileDialog()

OpenDialog.InitialDirectory = "D:\"

OpenDialog.FileName = ""

OpenDialog.DefaultExt = "jpg"

OpenDialog.Filter = "Jpg Files(*.jpg)|*.jpg|Gif Files(*.gif)|*.gif"

OpenDialog.CheckFileExists = True

OpenDialog.CheckPathExists = True

 

Dim ResultDialog As DialogResult = OpenDialog.ShowDialog()

If Not OpenDialog.FileName.EndsWith(".jpg") And Not OpenDialog.FileName.EndsWith(".gif") And Not _ OpenDialog.FileName.EndsWith(".GIF") And Not OpenDialog.FileName.EndsWith(".JPG") Then

MessageBox.Show("Please Upload JPEG & GIF Files ONLY", "At-Taiseer", MessageBoxButtons.OK)

ElseIf ResultDialog = DialogResult.OK Then

resumepic.Image = Image.FromFile(OpenDialog.FileName)

strfilename = OpenDialog.FileName

End If

dim frm as new form2

frm.show

me.hide

 

End Sub 'OpenPicDialog

 

 

 

'form2

Dim sr As StreamReader

Dim srmoze As String

Dim arrldb() As String

sr = New StreamReader("Ldb.txt")

srmoze = sr.ReadLine()

sr.Close()

 

 

thanx in advance

 

Mu$t@f@:confused: :confused:

  • *Experts*
Posted

As far as I know whenever you use the Dialog of that sort like OpenFileDialog, the location that you last used will be saved.

Use this as a path:

Application.StartupPath & "\ldb.txt"

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