Question

我有一个搜索和打开文件的代码：

def OpenButton(self, event):
    filedialog = wx.FileDialog(self, message = 'Open text file',
        defaultDir = '.',
        defaultFile = 'TestTOC.txt',
        wildcard = "Text source (*.txt)|*.txt|"  "All files (*.*)|*.*",
        style = wx.OPEN)
    if filedialog.ShowModal() == wx.ID_OK:
        print filedialog.GetPath()
    event.Skip()

它会显示文件的路径：C:\....\Desktop\test.txt

我有另一个代码需要读取我选择的文件：

def ReadButton(self, event):
     file=open('C:....\Desktop\test.txt','r')    # the same path as above
     text=file.read() 
     file.close()

如何复制该路径并将其替换为open（....，'r'）？

Answer 1

使用变量？

    def OpenButton(self, event):
        filedialog = wx.FileDialog(self, message = 'Open text file',
           defaultDir = '.',
            defaultFile = 'TestTOC.txt',
            wildcard = "Text source (*.txt)|*.txt|"  "All files (*.*)|*.*",
            style = wx.OPEN)
        if filedialog.ShowModal() == wx.ID_OK:
            self.filepath = filedialog.GetPath()
        event.Skip()

    def ReadButton(self, event):
         file=open(self.filepath,'r')    # the same path as above
         text=file.read() 
         file.close()

Answer 2

更改

print filedialog.GetPath()

到

path = filedialog.GetPath()
print path

然后用路径变量做任何你想做的事。

在python中重用一个路径

2 个答案: