Python tkFileDialog.asksaveasfile - 获取文件路径 [英] Python tkFileDialog.asksaveasfile - get file path

问题描述

我想获取文件exportFile"的路径.

I want to get path of file "exportFile".

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

如果我写打印导出文件"，我得到:

If I write "print exportFile", I get:

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

但我只需要路径 - C:/Users/Desktop/Test/aaaa.txt".有什么解决办法吗?谢谢.

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

推荐答案

试试这个:

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

它会返回:

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

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