将目录名称传递给askopenfilename

Question

我有三个功能；用户选择dirBut一个目录，其输出进入dirname并更新一个Entry框。在第三个函数 中，dataInput用户选择一个文件。我希望打开文件对话框在用户先前选择并由 定义的目录中打开dirname，但是，我不确定如何传递dirname给句柄，以便我可以使用它，因为askopenfilename是askdirectory从按钮调用的。

def UserFileInput(self,status,name):
    row = self.row
    optionLabel = tk.Label(self)
    optionLabel.grid(row=row, column=0, sticky='w')
    optionLabel["text"] = name
    text = status
    var = tk.StringVar(root)
    var.set(text)
    w = tk.Entry(self, textvariable= var)
    w.grid(row=row, column=1, sticky='ew')
    self.row += 1

    return w, var


def askdirectory(self):
    dirname = tkFileDialog.askdirectory()
    if dirname:
      self.directoryEntry.delete(0, tk.END)
      self.directoryEntry.insert(0, dirname)


def askfilename(self):
  filename = tkFileDialog.askopenfilename(initialdir=dirname)
  if filename:
    self.dataInput.delete(0, tk.END)
    self.dataInput.insert(0, filename)

currentDirectory = os.getcwd()
directory,var = self.UserFileInput(currentDirectory, "Directory")
self.directoryEntry = directory

dirBut = tk.Button(self, text='Select directory...', command = self.askdirectory)
dirBut.grid(row=self.row-1, column=2)

dataInput, var = self.UserFileInput("", "Data input")
self.dataInput = dataInput

fileBut = tk.Button(self, text='Select input file...', command = self.askfilename)
fileBut.grid(row=self.row-1, column=2)

Answer 1

假设askdirectory和askfilename属于同一类，请尝试将目录分配给self.dirname而不是dirname. 然后该变量将在类中的任何位置可见。

def askdirectory(self):
    self.dirname = tkFileDialog.askdirectory()
    if self.dirname:
      self.directoryEntry.delete(0, tk.END)
      self.directoryEntry.insert(0, self.dirname)


def askfilename(self):
  filename = tkFileDialog.askopenfilename(initialdir=self.dirname)
  if filename:
    self.dataInput.delete(0, tk.END)
    self.dataInput.insert(0, filename)