Question

我有一个gui应用程序，有几个窗口和按钮前进和后退。为此，我使用控制器并在每次更改窗口时将帧提升到顶部。这是我的控制器代码和典型的框架。

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视图，但我想知道是否有可能在每次引发帧时都运行一个函数，类似于Android中的onResume函数。我希望这样做，以防一些基础数据像这个上传页面一样发生了变化。例如，如果从填充listview的数组中删除某个项目，我希望能够刷新它。

Answer 1

Tkinter有低级事件，例如<Visibility>和<Map>，当页面发生变化时会触发这些事件。不幸的是，这些并不能在所有平台上可靠地运行。

最简单，最强大的解决方案是生成您自己的事件。您可以通过在<<和>>中指定事件来创建并绑定到自定义事件（例如：<<ShowFrame>>）。

首先，更改show_frame以在显示事件时将事件发送到窗口：

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

接下来，如果需要在可见的情况下通知它，每个页面都可以绑定到此事件：

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...")

如何在每次在tkinter中显示帧时运行一个方法

1 个答案: