如何在不调用python中的函数的情况下从外部访问函数内部声明的列表

Question

iam尝试使用tkinter-askopenfile((方法指定的位置使用"打开"按钮打开文件，如下所示：

def open_file():
file = askopenfile(mode ='r', filetypes =[('text files', '*.txt')])
if file is not None:
content = file.readlines()
btn = Button(root, text ='Open', command = lambda:open_file())

我想在不调用open_file((的情况下访问函数外的"content"，以避免重复打开窗口；

content = None
def open_file():
global content
file = askopenfile(mode ='r', filetypes =[('text files', '*.txt')])
if file is not None:
content = file.readlines()
return content
btn = Button(root, text ='Open', command = lambda:open_file())
print(content)

但我得到的输出是"无"，而不是文件内容。这个问题有什么解决办法吗？提前感谢

Answer 1

def open_file():
global content
file = askopenfile(mode ='r', filetypes =[('text files', '*.txt')])
if file is not None:
content = file.readlines()
print(content)
else:
content = ""
print(" ")
btn = Button(root, text ='Open', command=open_file).pack()