[Tutor] [wxPython-users] How to save file name of file opened fromwx.FileDialog ?

[Varsha Purohit]

Hi Alan,
    I am actually calling the binding function and then writing it
into the text value... i tried using simple print in the openfile
function and it shows the filename. I am trying to return the file
name value but even that is not responding...

class ScrolledWindow(wx.Frame):
    def __init__(self, parent, id, title):
        wx.Frame.__init__(self, parent, id, title, size=(350, 300))
        panel = wx.Panel(self, -1)
        txt1 = wx.TextCtrl(panel, -1, pos=(30, 100), size=(150, 20))
        button11 = wx.Button(panel, -1, "Open", pos=(200,100))
        name = self.Bind(wx.EVT_BUTTON, self.OnOpen, button11)
        txt1.write(name)

        self.Centre()
        self.Show()

    def OnOpen(self,event):
        self.dirname = ''
        dlg = wx.FileDialog(self, "Choose a file", self.dirname,"",
"*.*", wx.OPEN)
        if dlg.ShowModal()==wx.ID_OK:
            self.filename=dlg.GetFilename()
            Fname = self.filename
            self.dirname=dlg.GetDirectory()
            #f=open(os.path.join(self.dirname, self.filename),'r')
            #self.control.SetValue(f.read())
           # self.txt1.WriteText(Fname)
            f.close()

        dlg.Destroy()
        return Fname

app = wx.App()
ScrolledWindow(None, -1, 'Aliens')
app.MainLoop()

I donno the alternative to this now... :(

On Nov 17, 2007 3:35 PM, Alan Gauld <alan.gauld at btinternet.com> wrote:
>
> "Varsha Purohit" <varsha.purohit at gmail.com> wrote
>
> > later part of the program.  But i am not able to get the file name
> > and
> > display it in the text control. Here is the sample code.
> >
> > Fname = '' #Global variable to hold the file name.
>
> You don't need this since its stored in self.filename.
>
> > class ScrolledWindow(wx.Frame):
> >    def __init__(self, parent, id, title):
> >        txt1 = wx.TextCtrl(panel, -1, pos=(30, 100), size=(150, 20))
>
> If you want to write to it you need to store it in the object so this
> should be self.txt1
>
> >        txt1.write(Fname)
>
> You can't write the filename yet as it hasn't been fetched
>
> >
> >    def OnOpen(self,event):
> >        self.dirname = ''
> >        dlg = wx.FileDialog(self, "Choose a file", self.dirname,"",
> > "*.*", wx.OPEN)
> >        if dlg.ShowModal()==wx.ID_OK:
> >            self.filename=dlg.GetFilename()
>
> Here you get the filename but don't write it to the text control
>
> HTH,
>
> Alan G.
>
>
> _______________________________________________
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>



-- 
Varsha Purohit