在多個tkinterEntry小部件中使用退格鍵

我在我以前的專案中制作了 wordle（使用 tkinter），現在我不想使用一個條目來猜測，而是希望將 5 個條目鏈接在一起，就像下面的塊一樣作業：

在影像中，您可以在這些“鏈接”條目之間退格。我弄清楚了條目之間的輸入，但是如何洗掉它們？

from tkinter import *

root = Tk()
root.geometry('400x400')


def testlen():
    global textinentry1, textinentry2, textinentry3, textinentry4, textinentry5
    textinentry1= entry1.get()
    textinentry2= entry2.get()
    textinentry3= entry3.get()
    textinentry4= entry4.get()
    textinentry5= entry5.get()
    if len(textinentry1)  >1 :
        entry1.delete(0,END)
        entry1.insert(0,textinentry1[0])
        entry2.delete(0, END)
        entry2.insert(0,textinentry1[1] )
        entry2.focus_set()
    if len(textinentry2)  >1 :
        entry2.delete(0,END)
        entry2.insert(0,textinentry2[0])
        entry3.delete(0, END)
        entry3.insert(0,textinentry2[1] )
        entry3.focus_set()
    if len(textinentry3)  >1 :
        entry3.delete(0,END)
        entry3.insert(0,textinentry3[0])
        entry4.delete(0, END)
        entry4.insert(0,textinentry3[1] )
        entry4.focus_set()
    if len(textinentry4)  >1 :
        entry4.delete(0,END)
        entry4.insert(0,textinentry4[0])
        entry5.delete(0, END)
        entry5.insert(0,textinentry4[1] )
        entry5.focus_set()
    if len(textinentry5) > 1:
        entry5.delete(0,END)
        entry5.insert(0,textinentry5[0])


entry1 = Entry(root,width=3,  font = ('Georgia 18'), justify=CENTER)
entry1.grid(row=0, column=0)
entry2 = Entry(root, width=3,  font = ('Georgia 18'), justify=CENTER)
entry2.grid(row=0, column=1)
entry3 = Entry(root, width=3, font = ('Georgia 18'), justify=CENTER)
entry3.grid(row=0, column=2)
entry4 = Entry(root, width=3, font = ('Georgia 18'), justify=CENTER)
entry4.grid(row=0, column=3)
entry5 = Entry(root, width=3, font = ('Georgia 18'), justify=CENTER)
entry5.grid(row=0, column=4)

def loop():
    testlen()
    root.after(1,loop) # 1 is 1 millisecond. Here root.after method calls the loop
                       # function after 1 millisecond without crashing your code.

loop()

root.mainloop()

uj5u.com熱心網友回復：

首先，您必須獲得一些東西來注冊按鈕按下。為此，我匯入keyboard并用于if keyboard.is_pressed("backspace"):查看是否按下了退格鍵。

不幸的是，當一個人按下一個按鈕時，我們按住它幾毫秒，所以對于計算機，我們按下它不止一次。為了防止這種情況，我匯入time并發現使用time.sleep(.15)是最好的延遲，因為它阻止了我不小心按一次以上，而且我也可以按住它而不用代碼慢慢擦除每個字符。

這可以用一個回圈更好地濃縮，但我的獨立盒子的代碼是：

from tkinter import *
import keyboard, time

root = Tk()
root.geometry('400x400')


def testlen():
    global textinentry1, textinentry2, textinentry3, textinentry4, textinentry5, whathasfocus
    whathasfocus= root.focus_get()
    textinentry1= entry1.get()
    textinentry2= entry2.get()
    textinentry3= entry3.get()
    textinentry4= entry4.get()
    textinentry5= entry5.get()


    if whathasfocus == entry2 and len(textinentry2) == 0:
        if keyboard.is_pressed("backspace"):
            time.sleep(.15)
            entry1.delete(0,END)
            entry1.focus_set()
    elif whathasfocus == entry3 and len(textinentry3) == 0:
        if keyboard.is_pressed("backspace") :
            time.sleep(.15)
            entry2.delete(0,END)
            entry2.focus_set()
    elif whathasfocus == entry4 and len(textinentry4) == 0:
        if keyboard.is_pressed("backspace") :
            time.sleep(.15)
            entry3.delete(0,END)
            entry3.focus_set()
    elif whathasfocus == entry5 and len(textinentry5) == 0:
        if keyboard.is_pressed("backspace") :
            time.sleep(.15)
            entry4.delete(0,END)
            entry4.focus_set()
    
    if len(textinentry1)  >1 :
        entry1.delete(0,END)
        entry1.insert(0,textinentry1[0])
        entry2.delete(0, END)
        entry2.insert(0,textinentry1[1] )
        entry2.focus_set()
    if len(textinentry2)  >1 :
        entry2.delete(0,END)
        entry2.insert(0,textinentry2[0])
        entry3.delete(0, END)
        entry3.insert(0,textinentry2[1] )
        entry3.focus_set()
    if len(textinentry3)  >1 :
        entry3.delete(0,END)
        entry3.insert(0,textinentry3[0])
        entry4.delete(0, END)
        entry4.insert(0,textinentry3[1] )
        entry4.focus_set()
    if len(textinentry4)  >1 :
        entry4.delete(0,END)
        entry4.insert(0,textinentry4[0])
        entry5.delete(0, END)
        entry5.insert(0,textinentry4[1] )
        entry5.focus_set()
    if len(textinentry5) > 1:
        entry5.delete(0,END)
        entry5.insert(0,textinentry5[0])


entry1 = Entry(root,width=3,  font = ('Georgia 18'), justify=CENTER)
entry1.grid(row=0, column=0)
entry1.focus_set()
entry2 = Entry(root, width=3,  font = ('Georgia 18'), justify=CENTER)
entry2.grid(row=0, column=1)
entry3 = Entry(root, width=3, font = ('Georgia 18'), justify=CENTER)
entry3.grid(row=0, column=2)
entry4 = Entry(root, width=3, font = ('Georgia 18'), justify=CENTER)
entry4.grid(row=0, column=3)
entry5 = Entry(root, width=3, font = ('Georgia 18'), justify=CENTER)
entry5.grid(row=0, column=4)
def loop():
    testlen()
    root.after(1,loop) # 1 is 1 millisecond. Here root.after method call the loop function after 1 millisecond without crashing your code.

loop()
root.mainloop()

標籤：Python tkinter 键盘事件 tkinter-entry

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