Python 从Tkinter中的列表创建数据输入屏幕

我试图找出如何从列表中生成数据输入屏幕

我在这里创建了我想要的基本功能：

from tkinter import *

root = Tk()
root.title("example")
root.geometry("1200x900")

def clicked():
    data = (foo_entry.get(), 
        bar_entry.get())
    success_label = Label(root, text = "Entries Stored!").grid(row = 3, column = 2)
    values_stored_label = Label(root, text = str(data)).grid(row = 4, column = 2)

foo_label= Label(root ,text = "foo").grid(row = 0,column = 0)
bar_label= Label(root ,text = "bar").grid(row = 1,column = 0)

foo_entry = Entry(root) 
foo_entry.grid(row =  0, column = 1) 
bar_entry = Entry(root) 
bar_entry.grid(row =  1, column = 1) 
submit_button = Button(root,text = "Submit Entries", command = lambda: clicked())
submit_button.grid(row = 2, column = 2)



root.mainloop()

我现在要做的是创建一个脚本，当它提供一个名称列表时，它会创建相同的数据输入屏幕。因此：

entries = ['foo','bar']

我基本上想从前面的示例中生成相同的屏幕，它创建窗口、标签和网格，但检索输入数据会在

条目中抛出一个关键错误。get（）

。 代码如下：

entries = ['foo','bar']
from tkinter import *

root = Tk()
root.title("example")
root.geometry("1200x900")

def clicked():
    readout= ''
    for i in range(len(entries)):
        entryget = {entries[i]: entry[entries[i]].get()}
        readout += entryget[entries[i]]
    success_label = Label(root, text = "Entries Stored!")
    success_label.grid(row = len(entries)+2,column = 1)
    submit_label = Label(root, text = readout)
    submit_label.grid(row = len(entries)+3,column = 1)


for i in range(len(entries)):
    label = {entries[i]: Label(root ,text = entries[i])}
    labelgrid = {entries[i]:label[entries[i]].grid(row = i,column = 0)}
    entry = {entries[i]: Entry(root)}
    entrygrid = {entries[i]: entry[entries[i]].grid(row = i, column = 1)}
 
submit_button = Button(root,text = "Submit Entries", command = lambda: clicked())
submit_button.grid(row = len(entries)+1, column = 1)
root.mainloop()

弹出的错误是：

 File "examp2.py", line 11, in clicked
    entryget = {entries[i]: entry[entries[i]].get()}
KeyError: 'foo'

任何帮助都会很棒。如果有更有效的方法，请告诉我。

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提前感谢

结束。您缺少的部分是，您需要创建字典或列表来存储循环之前的条目，而不是循环中的条目。试试这个：

from tkinter import *

def clicked():
    results = [entry.get() for entry in entries.values()]
    readout= ', '.join(results)
    success_label = Label(root, text = "Entries Stored!")
    success_label.grid(column = 1) # row defaults to the next available one
    submit_label = Label(root, text = readout)
    submit_label.grid(column = 1)

LABELS = ['foo','bar']
entries = {} # initialize an empty dictionary to fill later

root = Tk()
root.title("example")
root.geometry("1200x900")

for i, text in enumerate(LABELS):
    label = Label(root, text = text) # no need to store the label
    label.grid(row = i,column = 0)
    entry = Entry(root)
    entry.grid(row = i, column = 1)
    entries[text] = entry # store this entry in the global dictionary

submit_button = Button(root,text = "Submit Entries", command = clicked) # don't need lambda 
submit_button.grid(column = 1)
root.mainloop()

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谢谢。这很好地解决了我的问题