Question

我在tkinter上遇到了这个问题。我想创建一个GUI来恢复通过askopenfilename选择的文件的路径和名称，然后将在后续代码中使用。我试过可能的选择，但我没有成功。我得到的最好的是跟随但不回报我需要的东西。谢谢你的帮助。

import tkinter as tk
from tkinter.filedialog import askopenfilename
class TkFileDialogExample(tk.Frame):

  def __init__(self, root):
    tk.Frame.__init__(self, root)
    self.a=[]
    tk.Button(self, text='askopenfilename', command=self.askopenfilename).pack()

  def askopenfilename(self):
      filename= askopenfilename()
      self.a.append(filename)
      return self.a

# MAIN PROGRAM
aa=[]
root = tk.Tk()
TkFileDialogExample(root).pack()
root.mainloop()
aa.append(TkFileDialogExample.askopenfilename)
print(aa)

Answer 1

我认为一个例子在评论中会有所帮助

import tkinter as tk
from tkinter.filedialog import askopenfilename

filenames = []

def open_file():

    filename = askopenfilename()
    if filename:
        filenames.append(filename)

root = tk.Tk()
tk.Button(root, text='Open File', command=open_file).pack()
root.mainloop()
print(filenames)

当您退出GUI时，您将获得一个列表，其中包含用户未单击取消的文件管理列表中所有有效打开的文件名。

Answer 2

下面的代码创建一个GUI，其上有一个按钮，弹出一个askopenfilename对话框并将结果添加到列表中。该按钮还向GUI添加一个标签，在其上显示askopenfilename对话框返回的文件路径。

import tkinter as tk
from tkinter.filedialog import askopenfilename

###Step 1: Create The App Frame
class AppFrame(tk.Frame):
    def __init__(self, parent, *args, **kwargs):
        ###call the parent constructor
        tk.Frame.__init__(self, parent, *args, **kwargs)

        self.parent = parent

        ###Create button
        btn = tk.Button(self, text='askopenfilename',command=self.askopenfilename)
        btn.pack(pady=5)

    def askopenfilename(self):
        ###ask filepath
        filepath = askopenfilename()

        ###if you selected a file path
        if filepath:
            ###add it to the filepath list
            self.parent.filepaths.append(filepath)

            ###put it on the screen
            lbl = tk.Label(self, text=filepath)
            lbl.pack()

###Step 2: Creating The App
class App(tk.Tk):
    def __init__(self, *args, **kwargs):
        ###call the parent constructor
        tk.Tk.__init__(self, *args, **kwargs)

        ###create filepath list
        self.filepaths = []

        ###show app frame
        self.appFrame = AppFrame(self)
        self.appFrame.pack(side="top",fill="both",expand=True)


###Step 3: Bootstrap the app
def main():
    app = App()
    app.mainloop()

if __name__ == '__main__':
    main()

在GUI中使用askopenfilename

2 个答案: