------ KEYBOARD ----------

Question

我正在使用Python v3.4并使用Tkinter库来训练我的GUI。

我在我的python应用程序中包含了一个屏幕键盘。问题如下：

他们是三个条目：

username = Entry(root,font="Verdana 22")

password = Entry(root, font="Verdana 22")

courrierEntry = Entry(root, font="Verdana 22")

下面是键盘

 def select(value):
        if value == "BACK":
            entry2 = len(username.get()[:-1])

        username.delete(entry2, END)

    elif value == " Space ":
            username.insert(END, ' ')

    elif value == " Tab ":
            username.insert(END, '    ')

    else:
            username.insert(END, value)

buttons = [
    'q','w','e','r','t','y','u','l','o','p','BACK','7','8','9','-',
    'a','s','d','f','g','h','j','k','l','[',']','4','5','6','+',
    'z', 'x','c','v','b','n','m',',','.','?','1','2','3','/','HELP',
    ]

varRow = 2
varColumn = 0

for button in buttons:

    command = lambda x = button: select(x)
    keyboardButton = Button(keyboardFrame, text = button, width = 8,  bg="#795548", fg="#ffffff",
            activebackground="#D7CCC8",activeforeground="#000000",relief = "raised", padx=0,font="Verdana 11 bold",
            pady=8, bd=8, command = command)
    keyboardButton.grid(row = varRow, column = varColumn)
    varColumn += 1
    if varColumn > 14 and varRow == 2:
        varColumn = 0
        varRow += 1
    if varColumn > 14 and varRow == 3:
        varColumn = 0
        varRow +=1

正如您在键盘上注意到的那样，它仅指向username entry，因此即使按下按钮时，只有username entry只有username entry时才会对#checking to see whcih input entry is currently focused inputFocus = None def currentEntry(userEntry): global inputFocus inputFocus = userEntry username.bind("<FocusIn>", lambda: currentEntry(username)) password.bind("<FocusIn>", lambda: currentEntry(password)) courrierEntry.bind("<FocusIn>", lambda: currentEntry(courrierEntry)) def select(value): if value == "BACK": entry2 = len(inputFocus.get()[:-1]) inputFocus.delete(entry2, END) elif value == " Space ": inputFocus.insert(END, ' ') elif value == " Tab ": inputFocus.insert(END, ' ') else: inputFocus.insert(END, value)没有注意力。

我希望建立某种if语句，问题是我不熟悉验证条目是否被集中/选中。它应该通过当前选择的条目而不是名称。因此，基本上每当我选择时我都可以输入。

更新

------ KEYBOARD ----------

class Organization < ActiveRecord::Base
  mount_uploader :image, ImageUploader
end

Answer 1

您可以调用keyboardFrame.focus_get()来获取具有焦点的窗口对象。

Answer 2

你想要的是记住最后一个有焦点的条目。因为按下按钮会将焦点对准按钮。

hasFocus = None

def currentEntry(myEntry):
    global hasFocus 
    hasFocus = myEntry

username.bind("<FocusIn>", lambda myEvent: currentEntry(username)) # do the same for the other entries

然后在select中使用hasFocus。

注意：您可能希望添加 if inputFocus != None:处于选择状态，因为在开头，没有选定的条目。

检查哪个条目是关注的（Python）

------ KEYBOARD ----------

2 个答案: