Question

以下代码是一个将按钮附加到现有程序的程序，因此可以在更友好的界面上而不是在代码内部进行选择。我试图使用下拉菜单，但setEthAnt1函数似乎有一个错误：TypeError：setEthAnt1（）不带参数（给定1）。我不知道什么是我无法传递的。有没有人有任何想法？

from Tkinter import *
import ThreegroupsGraphics as three

def run():
    three.main()

def setEthAnt1():
    name = var.get()
    print name
    three.OneTo2Ant = name
    print three.OneTo2Ant

root = Tk()
var = StringVar()
var.set("Group 1 Ethnic Antagonism")
OptionMenu(root, var, "1","2","3","4","5","6","7","8","9","10", command = setEthAnt1).pack()
butn = Button(root, text = 'run',  command = run)
butn.pack()
root.mainloop()

Answer 1

当您为OptionMenu指定命令时，所选项目的值将被发送到命令中，这实际上使您的var.get（）不需要。见下文：

from Tkinter import *
import ThreegroupsGraphics as three

def run():
    three.main()

def setEthAnt1(name):
    print name
    three.OneTo2Ant = name
    print three.OneTo2Ant

root = Tk()
var = StringVar()
var.set("Group 1 Ethnic Antagonism")
OptionMenu(root, var, "1","2","3","4","5","6","7","8","9","10", command = setEthAnt1).pack()
butn = Button(root, text = 'run',  command = run)
butn.pack()
root.mainloop()

如果您不希望setEthAnt1拥有任何参数并且仍然使用var.get()，您可以像OptionMenu那样为OptionMenu(root, var, "1","2","3","4","5","6","7","8","9","10", command = lambda _: setEthAnt1).pack() lamda函数创建命令：

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问题与python Tkinter下拉菜单

1 个答案: