我正在编写一个程序,该程序最终将要求用户输入文件名并图形化数据文件。但是,我很难将用户输入的文件名传输到下一页。
我已经实现了本页(How to access variables from different classes in tkinter?)上显示的两个答案,但是很遗憾,这个答案无法起作用。我在想也许PageOne正在被重写,或者entry_field_variable正在被重写或未保存。我是tkinter的新手,因此我可能会进行明显的疏忽,但我们将为您提供任何帮助。
import tkinter as tk
LARGE_FONT= ("Verdana", 12)
NORM_FONT= ("Verdana", 10)
SMALL_FONT= ("Verdana", 8)
class mGui(tk.Tk):
def __init__(self, *args, **kargs):
tk.Tk.__init__(self, *args, **kargs)
container = tk.Frame(self)
container.grid_rowconfigure(0, weight=1)
container.grid_columnconfigure(0, weight=1)
container.grid()
self.shared_data = {"entry_field_variable":tk.StringVar()}
self.frames = {}
for F in (StartPage, PageOne):
frame = F(container, self)
self.frames[F] = frame
frame.grid(row=0, column=0, sticky="nsew")
self.show_frame(StartPage)
def show_frame(self, cont):
frame = self.frames[cont]
frame.tkraise()
class StartPage(tk.Frame):
def __init__(self, parent, controller):
self.controller = controller
tk.Frame.__init__(self, parent)
label_1 = tk.Label(self, text = "Start Page", font = LARGE_FONT)
label_1.grid(row = 0, column = 0)
entry1 = tk.Entry(self, textvariable = self.controller.shared_data["entry_field_variable"])
entry1.grid(row = 1, column = 0)
button1 = tk.Button(self,text = "Go", command = lambda: controller.show_frame(PageOne))
button1.grid(row = 2, column = 0)
class PageOne(tk.Frame):
def __init__(self, parent, controller):
self.controller = controller
tk.Frame.__init__(self, parent)
label_1 = tk.Label(self, text = "Page One", font = LARGE_FONT)
label_1.grid(row = 1, column = 4)
filename = self.controller.shared_data["entry_field_variable"].get()
label_2 = tk.Label(self, text = filename, font = LARGE_FONT)
label_2.grid(row = 2, column = 0)
button1 = tk.Button(self, text = "Go Back", command = lambda: controller.show_frame(StartPage))
button1.grid(row = 4, column = 1)
app = mGui()
app.mainloop()
答案 0 :(得分:0)
您使用tk.StringVar
的想法很好,但是您缺少一些东西:
entry
的{{1}}字段中输入的值。 StartPage
必须有权访问它,因此有必要将保存文件名的PageOne
分配给标签tk.StringVar
。这是可以完成的方法:
textvariable