Python Tkinter/如何让OptionMenus共享一项列表？

Question

我正在尝试构建共享相同“基本项目列表”的多个选项菜单。不可能在不同菜单中多次选择一项，因此当在可用菜单之一中选择一项时，必须更新所有菜单。

from tkinter import *

# for example 5 fields
number_of_fields = 5
starting_list = ["item1","item2","item3","item4","item5"]
entry_list = []
option_list = []
option_var = []


def quit():
    raise SystemExit()

# if an item is selected in one of the
# menus run this function
def reset_menu(sel_item):
    # for each field
    for field in range(number_of_fields):
        new_list = []
        selection = option_var[field].get()
        # look for selected items in all menus
        # and build new list which contains all
        # items from the starting_list minus the
        # items which are already selected
        # keep the one selected (for a menu itself)
        for option in starting_list:
            marker = 0
            for j in range(number_of_fields):
                if(str(option_var[j].get()) == str(option)):
                    marker = 1
            if(marker == 0):
                new_list.append(str(option))
            else:
                pass
            if(str(selection) == str(option)):
                new_list.append(str(option))
        # print new generated item list
        # just to be sure it works so far
        print("field",field,"new list=",new_list)

        # NOW HERE SOMETHING IS WRONG I GUESS
        # empty menu
        option_list[field]["menu"].delete(0, "end")
        # add new menu items
        for item in new_list:
            option_list[field]['menu'].add_command(label=item, command=lambda value=item:option_var[field].set(value))


root = Tk()
root.title("OptionMenu")

# menu variable for each field
for i in range(number_of_fields):
    option_var.append(StringVar(root))

# initial value for each field 
for i in range(number_of_fields):
    option_var[i].set("")

# create menu for each field
for i in range(number_of_fields):
    option_list.append(OptionMenu(root, option_var[i], *starting_list, command=reset_menu))

# create entry for each field
for i in range(number_of_fields):
    entry_list.append(Entry(root))

# build gui
for i in range(number_of_fields):
    entry_list[i].grid(row=int(i),column=0,sticky=N+S+W+E)
    option_list[i].grid(row=int(i), column=1,sticky=N+S+W+E)
button = Button(root, text="OK", command=quit)
button.grid(row=number_of_fields,column=1,sticky=N+S+W+E)

mainloop()

现在一切似乎都很好，直到我尝试更新菜单。新的菜单项列表已正确生成(请参阅打印语句)，并且菜单具有正确的项目，但在选择一个菜单后，唯一更改其选定状态的菜单是最后一个菜单。有什么想法吗？

问候现货

Answer 1

我发现你的问题是因为我也在尝试完成同样的任务。在 dir(tkinter) 中进行了一番探索后，我找到了一个解决方案，您启发了我创建一个帐户来发布。

我在代码中保留了您的原始评论，而我未更改的部分。

首先，用于生成选项的代码不必要地困惑。从完整列表中删除项目似乎更干净，而不是手动从空列表中填充列表。

您当前正在使用 tkinter.OptionMenu()。如果您改为使用 tkinter.ttk.OptionMenu()，它有一个名为 set_menu(*values) 的方法，该方法接受任意数量的值作为其参数，并将该菜单的选项设置为这些参数。

如果您进行切换，需要注意一件事 - ttk 的 OptionMenu 不允许在下拉列表中选择其默认值，因此建议将该值设为空白，就像我在starting_list 的声明中所做的那样。

为了保留空白选项，我添加了一个额外的空白选项，以便可以选择它。这样，如果您错误地选择了错误的选项，您可以恢复您的选择。

from tkinter import *
from tkinter.ttk import *

# for example 5 fields
number_of_fields = 5
starting_list = ["","item1","item2","item3","item4","item5"]
entry_list = []
option_list = []
option_var = []


def quit():
    raise SystemExit()

# if an item is selected in one of the
# menus run this function
def reset_menu(sel_item):
    # for each field
    for field in range(number_of_fields):
        new_list = [x for x in starting_list]
        selection = option_var[field].get()
        # look for selected items in all menus
        # and build new list which contains all
        # items from the starting_list minus the
        # items which are already selected
        # keep the one selected (for a menu itself)
        for option in starting_list[1:6]:
            #add selectable blank if option is selected
            if (str(selection) == str(option)):
                    new_list.insert(0,"")
            for j in range(number_of_fields):
                if(str(selection) != str(option) and str(option_var[j].get()) == str(option)):
                    new_list.remove(option) 
        # print new generated item list
        # just to be sure it works so far
        print("field",field,"new list=",new_list)
        #set new options
        option_list[field].set_menu(*new_list)

root = Tk()
root.title("OptionMenu")

# menu variable for each field
for i in range(number_of_fields):
    option_var.append(StringVar(root))

# initial value for each field 
for i in range(number_of_fields):
    option_var[i].set("")

# create menu for each field
for i in range(number_of_fields):
    option_list.append(OptionMenu(root, option_var[i], *starting_list, command=reset_menu))

# create entry for each field
for i in range(number_of_fields):
    entry_list.append(Entry(root))

# build gui
for i in range(number_of_fields):
    entry_list[i].grid(row=int(i),column=0,sticky=N+S+W+E)
    option_list[i].grid(row=int(i), column=1,sticky=N+S+W+E)
button = Button(root, text="OK", command=quit)
button.grid(row=number_of_fields,column=1,sticky=N+S+W+E)

mainloop()

您可能想要研究的是如何提高您的期权生成效率。现在，对于 n 个选项，您将循环菜单 n^2 次。我建议考虑传递刚刚在回调中选择的值，而不是搜索每个菜单以查看之前选择的内容。

作为一个额外的小注意事项，您的“确定”按钮会导致崩溃。我不确定这是故意行为、我的系统怪癖还是其他原因。

希望这会有所帮助!