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OpenFileDialog


XpressMad
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eu queria buscar o caminho de uma imagem para depois utilizar o caminho

mas ao executar o programa quando carrego no butao que contem o codigo dame erro

este e o meu coodigo

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
        OpenFileDialog1.Title = "Selecione a imagem"
        OpenFileDialog1.InitialDirectory = "C:\imagens"
        OpenFileDialog1.ShowDialog()     'o erro é aqui'
    End Sub

    Private Sub OpenFileDialog1_FileOk(ByVal sender As System.Object, ByVal e As System.ComponentModel.CancelEventArgs) Handles OpenFileDialog1.FileOk

        Dim strm As System.IO.Stream
        strm = OpenFileDialog1.OpenFile()
        TextBox1.Text = OpenFileDialog1.FileName.ToString()

    End Sub

Desde ja muito obrigado

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Hmm, só para despiste, experimenta da seguinte forma:

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
    Dim OFD As New OpenFileDialog
    OFD.ShowDialog()
End Sub

Sérgio Ribeiro


"Great coders aren't born. They're compiled and released"
"Expert coders do not need a keyboard. They just throw magnets at the RAM chips"

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