دوست عزیز کد اقای یاوری را به این صورت تصحیح کنید مشکل حل می شود
Dim strPath As String = Application.StartupPath & "\Image"
Dim strFilePath As String = strPath & "\1.jpg"
Dim stream As IO.FileStream
Private Sub picTest_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles picTest.Click
If IO.File.Exists(strPath) Then
Else
If MessageBox.Show("آیا مایل به درج عکس هستید", "درج عکس", MessageBoxButtons.YesNo, MessageBoxIcon.Question) = Windows.Forms.DialogResult.Yes Then
If dlgOpen.ShowDialog() = Windows.Forms.DialogResult.OK Then
MsgBox(dlgOpen.FileName)
picTest.Image = Drawing.Image.FromFile(dlgOpen.FileName)
saveImage(dlgOpen.FileName)
picTest.Image = Drawing.Image.FromFile(dlgOpen.FileName)
End If
End If
End If
End Sub
Sub saveImage(ByVal strFileName As String)
'If IO.File.Exists(strFileName) Then
If Not IO.Directory.Exists(strPath) Then
IO.Directory.CreateDirectory(strPath)
End If
If IO.File.Exists(strFilePath) = True Then
picTest.Image = Nothing
stream.Close()
IO.File.Delete(strFilePath)
End If
IO.File.Copy(strFileName, strFilePath)
'End If
End Sub
Private Sub frmTestPic_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load
If IO.File.Exists(strFilePath) Then
stream = New IO.FileStream(strFilePath, IO.FileMode.Open)
picTest.Image = Drawing.Image.FromStream(stream)
End If
End Sub
این هم خود پروژه
TestAddPic.zip