我有一个列表和 3 个路径。每条路径都指的是专门为该国家/地区运行的应用程序。
country = ['Spain', 'United Kingdom', 'Malaysia']
path_spain = r"c:\data\FF\Desktop\PythonFolder\spain_software.py"
path_uk = r"c:\data\FF\Desktop\PythonFolder\uk_software.py"
path_malaysia = r"c:\data\FF\Desktop\PythonFolder\malaysia_software.py"
我创建的运行按钮
需要根据我在OptionMenu
中选择的国家/地区触发这3个应用程序之一。因此,如果我选择Malaysia
,我希望Run 按钮
在path_malaysia
中运行应用程序。我正在努力解决这个问题。如果我在 OptionMenu
中单击 Malaysia
,我最好还希望将 Run 按钮
更改为 Run application Malaysia
示例。
这是我的代码:
import os
from tkinter import *
window = Tk()
window.title("Running Python Script") #Create window
window.geometry('550x300') #geo of the window
def run():
os.system('python path_spain')
#The run button (this button runs some other software)
run_button = Button(window, text="Run application.....", bg="blue", fg="white",command=run)
run_button.grid(column=0, row=2)
#These are the option menus
dd_country = StringVar(window)
dd_country.set(country [0]) #the first value
w = OptionMenu(window, dd_country, *country)
w.grid(row=0,column=1)
#These are the titles
l1 = Label(window, text='Select Country', width=15 )
l1.grid(row=0,column=0)
mainloop()
目前仅适用于西类牙...
最佳答案
import os
from tkinter import *
owner = ['Spain', 'United Kingdom', 'Malaysia']
path_spain = r"c:\data\FF\Desktop\PythonFolder\spain_software.py"
path_uk = r"c:\data\FF\Desktop\PythonFolder\uk_software.py"
path_malaysia = r"c:\data\FF\Desktop\PythonFolder\malaysia_software.py"
window = Tk()
window.title("Running Python Script") # Create window
window.geometry('550x300') # geo of the window
def run():
if dd_owner.get() == "Spain":
print("spain")
# os.system('python path_spain')
elif dd_owner.get() == "United Kingdom":
os.system('python path_uk')
elif dd_owner.get() == "Malaysia":
os.system('python path_malaysia')
def update_button(_):
run_button.config(text="Run application {}".format(dd_owner.get()))
# The run button (this button runs some other software)
# These are the option menus
dd_owner = StringVar(window)
dd_owner.set(owner[0]) # the first value
w = OptionMenu(window, dd_owner, *owner, command=update_button)
# w.config()
w.grid(row=0, column=1)
run_button = Button(window, text="Run application {}".format(dd_owner.get()), bg="blue", fg="white",command=run)
run_button.grid(column=0, row=2)
# These are the titles
l1 = Label(window, text='Select Owner', width=15)
l1.grid(row=0, column=0)
mainloop()
这将解决您的问题
关于python - 如何根据 tkinter 菜单中选定的选项触发按钮?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/67701045/