Question

我正在尝试使用按钮进行一些输入，但我必须使用Key来首先进行一个interraction to entry widget。我想在没有它的情况下实现它，但问题是当你进行多次输入时有没有办法用mause click和cursor选择那个条目？这是我的代码。

# -*- coding: utf-8 -*-
from Tkinter import*
import sys

pencere = Tk()
#pencere.overrideredirect(1)#başlıksız pencere
boyut = pencere.geometry ("400x400+250+150")#x*y +x+y boyut ve konum
baslik = pencere.title("guc hesaplama programi") #pencere başlığı
say=0
def click(key):
   global  say
   if say==0:
      if key == 'C':
        joule_entry.delete(0, END)  # clear entry
    else:
        joule_entry.insert(END, key)
    if say==1:
       if key == 'C':
        time_entry.delete(0, END)  # clear entry
    else:
        time_entry.insert(END, key)
    if say==2:
        if key == 'C':
        power_entry.delete(0, END)  # clear entry
    else:
        power_entry.insert(END, key)
def write_joule():
   global  say
   say=0

def write_time():
    global  say
    say=1

def write_power():
    global  say
    say=2

def joule_find():
    joule_entry.delete(0,END)
    power_t = power_entry.get()
    power = float(power_t)
    time_t = time_entry.get()
    time = float(time_t)

    joule = time*power/1000

    if 0 <= power <= 1500 and 0 <= time <= 9999:  #sınırlamalarımız

        joule_entry.insert(0,str(round(joule,3)))

    else:
        joule_entry.insert(0,"error")

def time_find():
    time_entry.delete(0,END)
    power_t = power_entry.get()
    power = float(power_t)
    joule_t = joule_entry.get()
    joule = float(joule_t)

    time = joule/(power/1000)

    if 0 <= power <= 1500 and 0 <= time <= 9999:  #sınırlamalarımız

        time_entry.insert(0,str(round(time,3)))

    else:
        time_entry.insert(0,"error")

def power_find():
   power_entry.delete(0,END)
   joule_t = joule_entry.get()
   joule = float(joule_t)
   time_t = time_entry.get()
   time = float(time_t)

   guc = 1000*joule/time

   if 0 <= guc <= 1500 and 0 <= time <= 9999:  #sınırlamalarımız

       power_entry.insert(0,str(round(guc,3)))

   else:
       power_entry.insert(0,"error")

joule_button = Button(text="ENERGY\t:" ,font="arial 14 bold", command =     joule_find)    #
joule_button.grid(row=0, column=0, sticky="w", pady=10)         

joule_entry = Entry()
joule_entry.grid(row=0, column=1, pady=10,sticky="news")

joule_button2 = Button(text="key" ,font="arial 14 bold", command = write_joule)    #
joule_button2.grid(row=0, column=3, sticky="w", pady=10)

joule_label = Label(text="Joule" ,font="arial 14 bold")
joule_label.grid(row=0, column=2, sticky="w", pady=10)

time_button = Button(text="TIME\t:",font="arial 14 bold", command = time_find )
time_button.grid(row=1, column=0, sticky="w", pady=10)

time_entry= Entry()
time_entry.grid(row=1, column=1, pady=10,sticky="news")

time_button2 = Button(text="key",font="arial 14 bold", command = write_time )
time_button2.grid(row=1, column=3, sticky="w", pady=10)

time_label= Label(text="Second",font="arial 14 bold" )
time_label.grid(row=1, column=2, sticky="w")

power_button = Button(text="Power\t:",font="arial 14 bold", command = power_find )
power_button.grid(row=2, column=0, sticky="w", pady=10)

power_entry= Entry()
power_entry.grid(row=2, column=1, pady=10,sticky="news")

power_button2 = Button(text="key",font="arial 14 bold", command = write_power )
power_button2.grid(row=2, column=3, sticky="w", pady=10)

power_label= Label(text="milli Watt",font="arial 14 bold" )
power_label.grid(row=2, column=2, sticky="w")

close = Button(text = "CLOSE",command = pencere.destroy)
close.grid(row=40,column=2,sticky="news")





btn_list = [
'7',  '8',  '9',
'4',  '5',  '6',
'1',  '2',  '3',
'0',  '.',  'C',]

# create all buttons with a loop
r = 5
c = 0

for b in btn_list:
   rel = 'ridge'
   cmd = lambda x=b: click(x)
   Button(text=b,width=5,relief=rel,command=cmd).grid(row=r,column=c,sticky="news")


c += 1
if c > 2:
    c = 0
    r += 1

mainloop()

python多个条目与屏幕键盘

0 个答案: