c++ - wxpython文件打开对话框

Question

为什么这个 wxpython 代码给我以下错误？

self.Bind(wx.EVT_MENU,self.onNewFile,self.New_File)

def onNewFile(self,evt):

    wx.FileDialog(None,'Choose a file',os.getcwd(),"",wx.OPEN)
    if dialog.ShowModal() == wx.ID_OK:
        print dialog.GetPath()
    dialog.Destroy()

还有其他设置菜单栏和创建项目的代码，但是当执行此代码时，出现以下错误:

Traceback (most recent call last):
  File "C:\Python27\Front_End.py", line 52, in onNewFile
    wx.FileDialog(None,'Choose a file',os.getcwd(),"",wx.OPEN)
  File "C:\Python27\lib\site-packages\wx-2.8-msw-unicode\wx\_windows.py", line 2430, in __init__
    _windows_.FileDialog_swiginit(self,_windows_.new_FileDialog(*args, **kwargs))
TypeError: String or Unicode type required

这是什么意思？

Answer 1

wx.FileDialog原型(prototype)如下

__init__(self, parent, id, title, pos, size, style, name)

您可能会遗漏一个参数。我还对您的代码进行了一些编辑，如下所示。

def onNewFile(self,evt):

    dialog = wx.FileDialog(None,'Choose a file',os.getcwd(),"", "",wx.OPEN)
    if dialog.ShowModal() == wx.ID_OK:
        print dialog.GetPath()
    dialog.Destroy()