Python tkFileDialog.asksaveasfile - get file path

Question

I want to get path of file "exportFile".

exportFile = tkFileDialog.asksaveasfile(mode='a')

If I write "print exportFile", I get:

<open file u'C:/Users/Desktop/Test/aaaa.txt', mode 'a' at 0x02CB6078>

But I need only path - "C:/Users/Desktop/Test/aaaa.txt". Is there any solution? Thank you.

Answer 1

Try this:

exportFile = tkFileDialog.asksaveasfile(mode='a')
exportFile.name

It'll return:

'C:/Users/Desktop/Test/aaaa.txt'

Answer 2

Use tkFileDialog.asksaveasfilename instead of tkFileDialog.asksaveasfile.

NOTE tkFileDialog.asksaveasfilename does not take mode parameter.

Answer 3

Try tkFileDialog.askdirectory instead of any file name dialog. That will return a directory instead of a file name.

Answer 4

Instead of printing 'exportFile' try to print 'exportFile.name'. It should give the output you desire