python - 我将如何制作一个每次在 tkinter 中显示帧时运行的方法

Question

我有一个 gui 应用程序，它有几个窗口和按钮可以前进和后退。为此，我使用 Controller 并在每次窗口更改时将框架提升到顶部。这是我的 Controller 代码和一个典型的框架。

import Tkinter as tk   # python
from tkFileDialog import askopenfilename, asksaveasfilename
from analyzer import Options

TITLE_FONT = ("Helvetica", 18, "bold")

class TennisProgram(tk.Tk):

    def __init__(self, *args, **kwargs):
        tk.Tk.__init__(self, *args, **kwargs)

        # the container is where we'll stack a bunch of frames
        # on top of each other, then the one we want visible
        # will be raised above the others
        container = tk.Frame(self)
        #Allow the window to be resized
        # container.resizable(width=True, height=True)
        container.pack(side="top", fill="both", expand=True)
        container.grid_rowconfigure(0, weight=1)
        container.grid_columnconfigure(0, weight=1)
        #List of options bundles. One bundle for each video
        self.options_bundle_list = {}
        self.frames = {}
        #Init empty array to hold the list of files
        #Placed in the container so that all views can access it
        self.files = []

        for F in (UploadPage, AnalysisOptionsPage, FormAnalysisOptions, MatchAnalysisOptions, MatchAnalysisOptionsPage2, OutputPage, ProcessingPage):
            page_name = F.__name__
            frame = F(container, self)
            self.frames[page_name] = frame

            # put all of the pages in the same location;
            # the one on the top of the stacking order
            # will be the one that is visible.
            frame.grid(row=0, column=0, sticky="nsew")
        # Name the window
        self.title("Tennis Analyzer")
        self.show_frame("UploadPage")

    def show_frame(self, page_name):
        '''Show a frame for the given page name'''
        frame = self.frames[page_name]
        frame.tkraise()


class UploadPage(tk.Frame):
#TODO Add logic to remove a file path from the list
    def __init__(self, parent, controller):
        tk.Frame.__init__(self, parent)
        self.controller = controller
        self.createView()


    def createView(self):
        label = tk.Label(self, text="Upload Videos for Analysis", font=TITLE_FONT)
        label.pack(side="top", fill="x", pady=10)

        #Listbox to display the list of files to be processed
        self.path_list = tk.Listbox(self)
        #Additional options allow listbox to expand as the window is resized
        self.path_list.pack(fill=tk.BOTH ,expand=True)

        for path in self.controller.files:
            self.path_list.insert(tk.END, path)

        add_vidoes_button = tk.Button(self, text="Add Videos",
                            command=self.choose_files)
        continue_button = tk.Button(self, text="Continue",
                            command=lambda: self.controller.show_frame("AnalysisOptionsPage"))
        add_vidoes_button.pack(side=tk.TOP)
        continue_button.pack(side=tk.BOTTOM)

    def choose_files_paths(self):
        #don't want a full GUI, so keep the root window from appearing
        #self.controller.withdraw()
        #Get a file name from the user
        filename = askopenfilename()
        #Add it to the list of files to be processed
        self.controller.files.append(filename)
        self.path_list.insert(tk.END, filename)

我在 init 中编写的代码执行一次并创建 View ，但我想知道是否有可能有一个函数在每次引发帧时运行，类似于 Android 中的 onResume 函数。我想这样做，以防一些基础数据发生变化，就像这个上传页面中那样。例如，如果从填充 ListView 的数组中删除了一个项目，我希望能够刷新它。

Answer 1

Tkinter 有低级事件，例如 <Visibility>和 <Map>当页面更改时应该触发。不幸的是，这些并不是在所有平台上都能可靠地工作。

最简单、最可靠的解决方案是生成您自己的事件。您可以通过在 << 中指定事件来创建并绑定(bind)到自定义事件和 >> (例如:<<ShowFrame>>)。

首先，更改show_frame在窗口显示时将事件发送到窗口:

def show_frame(self, page_name):
    ...
    frame.event_generate("<<ShowFrame>>")

接下来，如果每个页面需要在可见时得到通知，可以绑定(bind)到这个事件:

class UploadPage(tk.Frame):
    def __init__(self, parent, controller):
        ...
        self.bind("<<ShowFrame>>", self.on_show_frame)

    def on_show_frame(self, event):
        print("I am being shown...")