python 在循环内获取标签的特定值，并通过循环外的菜单获取它

Question

问题的简单示例

# import * is bad, this is just an example
from tkinter import *

root = Tk()
root.minsize(200, 200)

dict = {"1": ("banane 1", "fresh"), "2": ("banane 2", "not fresh"), "3": ("banane 3", "rotten")}

# Right click label
def openMenuGroup(self):
   menuGroup.post(self.x_root, self.y_root)
           
def closeMenuGroup():
    menuGroup.unpost()
           
menuGroup = Menu(root, tearoff=0)

for key, value in dict.items():
    name, quality = value
    lab = Label(text=name)
    lab.quality = quality
    lab.bind("<Button-3>", openMenuGroup)
    lab.pack()

menuGroup.add_command(label="Check quality", command=lambda:checkQuality(lab.quality))
menuGroup.add_command(label="Close", command=closeMenuGroup())
        
def checkQuality(self):
   print(self)
   
mainloop()

当您单击“检查质量”时，它将始终返回“烂”（最后一次迭代）。如何为每个标签获得正确的 lab.quality？

Answer 1

您需要在

openMenuGroup()

内创建弹出菜单，并将正确的质量传递给

checkQuality()

:

from tkinter import *

root = Tk()
root.minsize(200, 200)

dict = {"1": ("banane 1", "fresh"), "2": ("banane 2", "not fresh"), "3": ("banane 3", "rotten")}

def checkQuality(quality):
    print(quality)

# Right click label
def openMenuGroup(event):
    # create menu here
    menuGroup = Menu(root, tearoff=0)
    # use event.widget to reference the label that trigger this function
    menuGroup.add_command(label="Check quality", command=lambda:checkQuality(event.widget.quality))
    menuGroup.add_command(label="Close", command=menuGroup.unpost)
    menuGroup.post(event.x_root, event.y_root)

for key, value in dict.items():
    name, quality = value
    lab = Label(text=name)
    lab.quality = quality
    lab.bind("<Button-3>", openMenuGroup)
    lab.pack()

mainloop()