Question

我特别迟钝。我正在迭代技术意大利语单词列表，并希望使用tkinter接口插入翻译。没有GUI就没有问题：我的问题是我无法弄清楚如何进行迭代，将单词加载到ttk.Label并等待ttk.Entry字段中的用户条目。我搜索并找到了解释，但我不知道如何应用这些建议。这是我的代码使用一个简单的单词列表：

from tkinter import ttk
import tkinter as tk

def formd():

    list_of_terms = ['aardvark', 'ant','zombie', 'cat', 'dog', 'buffalo','eagle', 'owl','caterpiller', 'zebra',  'orchid','arum lily' ]
    discard_list = []
    temp_dict={}
    list_of_terms.sort()

    for item in list_of_terms:
        listKey.set(item)

    # need to wait and get user input
        translation =dictValue.get()
        temp_dict[item]=translation
        discard_list.append(item)
    # check if it has worked
    for key, value in temp_dict.items():
        print(key, value) 
# GUI for dict from list
LARGE_FONT= ("Comic sans MS", 12)

root = tk.Tk()
root.title('Nautical Term Bilingual Dictionary')
ttk.Style().configure("mybtn.TButton", font = ('Comic sans MS','12'), padding = 1,  foreground = 'DodgerBlue4')
ttk.Style().configure('red.TButton', foreground='red', padding=6, font=('Comic sans MS',' 10'))
ttk.Style().configure('action.TLabelframe', foreground = 'dodger blue3')
#.......contents frames.....................
nb = ttk.Notebook(root)
page5 = ttk.Frame(nb)
# declare variables
listKey= tk.StringVar()
dictValue = tk.StringVar()
# widgets
keyLabel =ttk.Label( page5, text = "key from list", font=LARGE_FONT).grid(row=3, column = 0)
Keyfromlist =ttk.Label(page5, width = 10, textvariable = listKey).grid(row = 3, column = 1)

valueLabel =ttk.Label( page5, text = "enter translation", font=LARGE_FONT).grid(row=3, column = 2)
listValue =ttk.Entry(page5, textvariable =dictValue, width = 15).grid(row = 3, column = 3)
#listValue.delete(0,'end')
#listValue.focus_set()

# add buttons
b1 = ttk.Button(page5, text="add to dictionary",style = "mybtn.TButton", command = formd)
b1.grid(row = 5, column = 0)
b5 = ttk.Button(page5, text="clear entry", style ="mybtn.TButton")
b5.grid(row = 5, column = 2)
nb.add(page5, text='From List')

nb.pack(expand=1, fill="both")   

for child in root.winfo_children():
    child.grid_configure(padx =5, pady=5)

if __name__ == "__main__":
    root.mainloop()

我想知道是否有人可以花时间提出解决方案。 How to stop a while loop to get input from user using Tkinter?是我无法想象如何在我的示例中使用的一个建议

Answer 1

tkinter不会＆＃34;在while循环中玩得很好＆＃34; 。

幸运的是，你不需要使用它。

您可以执行以下操作：

from tkinter import *

class App:
    def __init__(self, root):
        self.root = root
        self.list = ['aardvark', 'ant','zombie', 'cat', 'dog', 'buffalo','eagle', 'owl','caterpiller', 'zebra',  'orchid','arum lily' ]
        self.text = Label(self.root, text="aardvark")
        self.entry = Entry(self.root)
        self.button = Button(self.root, text="Ok", command=self.command)
        self.text.pack()
        self.entry.pack()
        self.button.pack()
    def command(self):
        print(self.text.cget("text")+" - "+self.entry.get())
        try:
            self.text.configure(text = self.list[self.list.index(self.text.cget("text"))+1])
        except IndexError:
            self.entry.destroy()
            self.button.destroy()
            self.text.configure(text = "You have completed the test")

root = Tk()
App(root)
root.mainloop()

这实质上使用Button小部件迭代到下一个文本并获得下一个输入。

迭代列表并使用tkinter GUI获取用户对每个项目的响应

1 个答案: