需要一个用于触摸屏的Gui键盘，该键盘在代码正确时会输出一个引脚 [英] Need a Gui Keypad for a touchscreen that outputs a pin when code is correct

问题描述

我有一个带有树莓派的树莓派，触摸屏上运行着树莓派，我希望在触摸屏上有一个Gui，上面有一个数字小键盘，当输入正确的输入时，一个大头针会输出到门闩或其他东西.我已经结束了用Python编写带有数字的Gui，但是我不能让几个数字彼此相邻.任何信息将对此有所帮助:) 这是我用来尝试放置按钮的代码(您可以看到我只是使用了一个简单的LED开/关按钮Gui，并使用它来查看按钮的位置)

I have a raspberry pi with a touchscreen running raspbian, I'm hoping to have a Gui on the touchscreen that had a number keypad that when a correct input is entered a pin will output to a door latch or something. I have been over to make a Gui with a number on (by Python) it but i cant get several numbers to sit next to each other. any info will help on this thanks :) This is the code I used to try and place the buttons (you can see i just used a simple LED on/off button Gui and used it to see the placement of the buttons)

from Tkinter import *
import tkFont
import RPi.GPIO as GPIO

GPIO.setmode(GPIO.BOARD)
GPIO.setup(40, GPIO.OUT)
GPIO.output(40, GPIO.LOW)

win = Tk()

myFont = tkFont.Font(family = 'Helvetica', size = 36, weight = 'bold')

def ledON():
    print("LED button pressed")
    if GPIO.input(40) :
        GPIO.output(40,GPIO.LOW)
                ledButton["text"] = "LED OFF"
    else:
        GPIO.output(40,GPIO.HIGH)
                ledButton["text"] = "LED ON"

def exitProgram():
    print("Exit Button pressed")
       GPIO.cleanup()
    win.quit()  


win.title("LED GUI")


exitButton  = Button(win, text = "1", font = myFont, command = ledON, height     =2 , width = 8) 
exitButton.pack(side = LEFT, anchor=NW, expand=YES)

ledButton = Button(win, text = "2", font = myFont, command = ledON, height = 2, width =8 )
ledButton.pack(side = TOP, anchor=CENTER, expand=YES)

ledButton = Button(win, text = "3", font = myFont, command = ledON, height = 2, width =8 )
ledButton.pack(side = RIGHT, anchor=NE, expand=YES)

ledButton = Button(win, text = "4", font = myFont, command = ledON, height = 2, width =8 )
ledButton.pack(side = TOP, anchor=W, expand=YES)

ledButton = Button(win, text = "5", font = myFont, command = ledON, height = 2, width =8 )
ledButton.pack(side = TOP, anchor=W, expand=YES)

ledButton = Button(win, text = "6", font = myFont, command = ledON, height = 2, width =8 )
ledButton.pack(side = TOP, anchor=W, expand=YES)

ledButton = Button(win, text = "7", font = myFont, command = ledON, height = 2, width =8 )
ledButton.pack(side = TOP, anchor=W, expand=YES)

ledButton = Button(win, text = "8", font = myFont, command = ledON, height = 2, width =8 )
ledButton.pack(side = TOP, anchor=W, expand=YES)

ledButton = Button(win, text = "9", font = myFont, command = ledON, height = 2, width =8 )
ledButton.pack(side = TOP, anchor=N, expand=YES)

ledButton = Button(win, text = "0", font = myFont, command = ledON, height = 2, width =8 )
ledButton.pack(side = TOP, anchor=NW, expand=YES)



mainloop()

推荐答案

带键盘的简单示例:

我使用全局字符串变量pin来保留所有按下的数字.
(您可以使用list代替string)

I use global string variable pin to keep all pressed numbers.
(You can use list instead of string)

键*删除最后一个数字，键#将pin与文本"3529"

Key * removes last number, key # compares pin with text "3529"

import tkinter as tk

# --- functions ---

def code(value):

    # inform function to use external/global variable
    global pin

    if value == '*':
        # remove last number from `pin`
        pin = pin[:-1]
        # remove all from `entry` and put new `pin`
        e.delete('0', 'end')
        e.insert('end', pin)

    elif value == '#':
        # check pin

        if pin == "3529":
            print("PIN OK")
        else:
            print("PIN ERROR!", pin)
            # clear `pin`
            pin = ''
            # clear `entry`
            e.delete('0', 'end')

    else:
        # add number to pin
        pin += value
        # add number to `entry`
        e.insert('end', value)

    print("Current:", pin)

# --- main ---

keys = [
    ['1', '2', '3'],    
    ['4', '5', '6'],    
    ['7', '8', '9'],    
    ['*', '9', '#'],    
]

# create global variable for pin
pin = '' # empty string

root = tk.Tk()

# place to display pin
e = tk.Entry(root)
e.grid(row=0, column=0, columnspan=3, ipady=5)

# create buttons using `keys`
for y, row in enumerate(keys, 1):
    for x, key in enumerate(row):
        # `lambda` inside `for` has to use `val=key:code(val)` 
        # instead of direct `code(key)`
        b = tk.Button(root, text=key, command=lambda val=key:code(val))
        b.grid(row=y, column=x, ipadx=10, ipady=10)

root.mainloop()

GitHub: furas/python-examples/tkinter/__button__/button-keypad

GitHub: furas/python-examples/tkinter/__button__/button-keypad

(我将链接更改为GitHub，因为我将代码移至了子文件夹)

( I changed link to GitHub because I moved code to subfolder)

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