Python Tkinter做所有條目滿時的東西

Question

我想在每個條目框在tk.Toplevel實例中已滿時運行命令。

目前，我擁有它，所以每次輸入新的輸入時，它都會檢查它們是否全部已滿。但是，這會變得效率低下，因爲如果輸入一個長字符串，它將檢查每個字母（因爲每次更改條目時都會執行一個條目的command）。另外，我還有大量的輸入框（總共大約30個）將被使用，所以這將變得非常低效。

are_EntriesFull = lambda master: all(bool(widget.get()) for widget in filter(lambda s: isinstance(s, tk.Entry), findAllWidgets(master))) 

_packingWidgets = (tk.Tk, tk.Toplevel, tk.Frame) 

def findAllWidgets(master, widgetsFound = None): 
    '''Returns all of the widgets in a packing instance''' 
    if widgetsFound is None: 
     widgetsFound = [] 
    subWidgets = master.grid_slaves() 
    for widget in subWidgets: 
     widgetsFound.append(widget) 
     if isinstance(widget, _packingWidgets): 
      widgetsFound.extend(findAllWidgets(widget, widgetsFound)) 
    return widgetsFound 

class GetWord: 

    '''Window to get input from the user and create an entry for the word''' 
     def __init__(self, master, current, total): 
      self.master, self.current, self.total, self.mainFrame = master, current, total, tk.Frame(master) 
      self.master.title(STANDARD_TITLE) 
      self.wordFrame, self.typeFrame, self.infoFrame, self.buttonFrame = tk.Frame(self.mainFrame), tk.Frame(self.mainFrame), tk.Frame(self.mainFrame), tk.Frame(self.mainFrame) 
      self.word = tk.StringVar(self.wordFrame) 
      self.wordLabel = tk.Label(self.wordFrame, text = "{}/{}) Word:".format(self.current, self.total), justify = tk.CENTER) 
      self.activateButton = lambda: self.okButton.configure(state = tk.ACTIVE) if are_EntriesFull(self.master) else None 
      self.wordEntry = tk.Entry(self.wordFrame, width = 50, justify = tk.CENTER, textvariable = self.word, command = self.activateButton) 
        # bunch of other code 

基本上什麼命令應該做的是激活按鈕（我想這隻有當每個條目被填充使用）。或者，我可以檢查按鈕被按下時的輸入框（並且始終保持激活狀態），並且只有當它們已滿時才執行按鈕的命令，但是我只想在滿足時激活它。 有沒有更好的方法來完成這個？謝謝！

Answer 1

這並不像您認爲的那樣低效。嘗試測量需要多長時間檢查30個（甚至300個或3000個）條目窗口小部件的內容。你可能會感到驚訝。在您注意到延遲之前，您可能必須每分鐘輸入幾百個字。

對於3000個條目小部件，在我的Mac上，我可以在每個按鍵上約3毫秒檢查它們的全部長度。這，假設我有一個小部件列表來檢查。如果你遍歷所有小部件和他們的孩子，這將花費更長的時間，這是a）毫無意義的，b）可能仍然非常快。

Answer 2

保存findAllWidgets的結果並重新使用該結果，而不是每次調用findAllWidgets。

#are_EntriesFull = lambda master: all(bool(widget.get()) for widget in filter(lambda s: isinstance(s, tk.Entry), findAllWidgets(master))) 
are_EntriesFull = lambda entries: all(bool(widget.get()) for widget in entries) # <--- 

_packingWidgets = (tk.Tk, tk.Toplevel, tk.Frame) 

def findAllWidgets(master, widgetsFound = None): 
    '''Returns all of the widgets in a packing instance''' 
    if widgetsFound is None: 
     widgetsFound = [] 
    subWidgets = master.grid_slaves() 
    for widget in subWidgets: 
     widgetsFound.append(widget) 
     if isinstance(widget, _packingWidgets): 
      widgetsFound.extend(findAllWidgets(widget, widgetsFound)) 
    return widgetsFound 

class GetWord: 

    '''Window to get input from the user and create an entry for the word''' 
     def __init__(self, master, current, total): 
      self.master, self.current, self.total, self.mainFrame = master, current, total, tk.Frame(master) 
      self.master.title(STANDARD_TITLE) 
      self.wordFrame, self.typeFrame, self.infoFrame, self.buttonFrame = tk.Frame(self.mainFrame), tk.Frame(self.mainFrame), tk.Frame(self.mainFrame), tk.Frame(self.mainFrame) 
      self.word = tk.StringVar(self.wordFrame) 
      self.wordLabel = tk.Label(self.wordFrame, text = "{}/{}) Word:".format(self.current, self.total), justify = tk.CENTER) 
      self.activateButton = lambda: self.okButton.configure(state = tk.ACTIVE if are_EntriesFull(all_entries) else tk.DISABLED) # <--- 
      self.wordEntry = tk.Entry(self.wordFrame, width = 50, justify = tk.CENTER, textvariable = self.word, command = self.activateButton) 
        # bunch of other code 

      ... 

      all_entries = findAllWidgets(self.master) # <---