Python中的tkinter怎么做？

Question

我已经为在线上找到的日历样式编写了算法。我希望输入一年后单击按钮在tk窗口中打印算法。不幸的是，它不会。我仅使用此代码作为算法的测试实现来测试算法。该算法将通过输入“ Mer”和年份（例如“ 2020”）来打印到终端，但是该算法不会在与终端相同的网格中打印到tkinter。它根本不会打印到tkinter窗口，并且没有错误消息，所以我不确定下一步该怎么做。

def make_calendar(year, start_day):
    tup =   [('Early Winter', range(1, 37, + 1)),
            ('Mid Winter 37,', range(37, 1, + 1)),
            ('Late Winter   ', range(38, 71, + 1)),
            ('Leap Day      ', range(70, 69 + 1)),#tup[3] is this line i need to delete if not leap year
            ('Late Winter   ', range(71, 74, + 1)),   
            ('Early Spring  ', range(1, 37, + 1)),
            ('Mid Spring 37,', range(37, 1, +1)),
            ('Late Spring   ', range(38, 74, + 1)),
            ('Early Summer  ', range(1, 37, + 1)),
            ('Mid Summer 37,', range(37, 1, +1)),
            ('Late Summer   ', range(38, 74, + 1)),
            ('Early Autumn  ', range(1, 37, + 1)),
            ('Mid Autumn 37,', range(37, 1, +1)),
            ('Late Autumn   ', range(38, 74, + 1)),
            ('Early Fall    ', range(1, 37, + 1)),
            ('Mid Fall 37,  ', range(37, 1, +1)),
            ('Late Fall     ', range(38, 74, + 1))];
    week = ['Mer', 'Ven', 'Ear', 'Mar', 'Jup', 'Sat', 'Ura', 'Nep', 'Plu']
    start_pos = week.index(start_day)
    def is_leap(year): 
        if (not year%4 and year%100 or not year%400): # algorithm to determine leap year
            return True
    if not is_leap(year):
        del tup[3]
    for month, days in tup: 
    # Print month title
        print('{0} {1}'.format(month, year).center(20, ' '))
        # Print Day headings
        print(''.join(['{0:<3}'.format(w) for w in week]))
        # Add spacing for non-zero starting position
        print('{0:<3}'.format('')*start_pos, end='')

        for day in days:
            # Print day
            print('{0!s:<3}'.format(day), end='')
            start_pos += 1
            if start_pos == 9:
                # If start_pos == 9 (Sunday) start new line
                print()
                start_pos = 0 # Reset counter
        print('
')
start_day=(input('Enter start day of the year Mer,Ven,Ear,Mar,Jup,Sat,Ura,Nep,Plu'))
year=int(input('Enter Year'))       
make_calendar(year, start_day)

这里，我已经设计了一个tkinter窗口，算法应该将其打印到该窗口，但不会将其打印到tk窗口，并且我不会收到任何错误消息。因此，基本上该算法可以在终端中工作并打印，但是当编码为在tkinter窗口中工作时，它什么也不返回。如果算法无法打印到它，则我创建的窗口小部件将无用。我已经测试了算法，并且可以正常工作。我已经测试了tkinter按钮，它们似乎无法打印。我不确定如何解决此问题。

import tkinter as tk


    #varContent = inputentry.get() # get what's written in the inputentry entry widget
    #outputtext.delete('0', 'end-1c') # clear the outputtext text widget
    #outputtext.insert(varContent)  # print content
    #these functions may be relevant to the solution I cannot figure out their implementation in this code.

HEIGHT = 700
WIDTH = 800


root = tk.Tk() #APPLICATION IS BETWEEN THIS LINE DOWN
canvas = tk.Canvas(root, height=HEIGHT, width=WIDTH)
canvas.pack()

frame = tk.Frame(root, bg='#80c1ff', bd='5')
frame.place(relx=0.5, rely=0.1, relwidth=0.75, relheight=0.1, anchor='n')

button = tk.Button(frame, text="Display Calendar", command=lambda: make_calendar(entry.get()))
button.grid(row=0, column=2)#this makes the buton apear in a grid

label = tk.Label(frame, text="Enter the year")
label.grid(row=0, column=0)#this makes the label apear in a grid 

entry = tk.Entry(frame)
entry.grid(row=0, column=1)#this makes the entry apear in a grid

lower_frame = tk.Frame(root, bg='#80c1ff', bd='5')
lower_frame.place(relx=0.5, rely=0.25, relwidth=0.75, relheight=0.6, anchor='n')

lower_label = tk.Text(lower_frame)
lower_label.place(relwidth=1, relheight=1)#this makes the label apear in a grid 

year=entry.get

def make_calendar(year):
    tup =   [('Early Winter', range(1, 37, + 1)),
            ('Mid Winter 37,', range(37, 1, + 1)),
            ('Late Winter   ', range(38, 71, + 1)),
            ('Leap Day      ', range(70, 69 + 1)),#tup[3] is this line i need to delete if not leap year
            ('Late Winter   ', range(71, 74, + 1)),   
            ('Early Spring  ', range(1, 37, + 1)),
            ('Mid Spring 37,', range(37, 1, +1)),
            ('Late Spring   ', range(38, 74, + 1)),
            ('Early Summer  ', range(1, 37, + 1)),
            ('Mid Summer 37,', range(37, 1, +1)),
            ('Late Summer   ', range(38, 74, + 1)),
            ('Early Autumn  ', range(1, 37, + 1)),
            ('Mid Autumn 37,', range(37, 1, +1)),
            ('Late Autumn   ', range(38, 74, + 1)),
            ('Early Fall    ', range(1, 37, + 1)),
            ('Mid Fall 37,  ', range(37, 1, +1)),
            ('Late Fall     ', range(38, 74, + 1))];
    week = ['Mer  ', 'Ven  ', 'Ear  ', 'Mar  ', 'Jup  ', 'Sat  ', 'Ura  ', 'Nep  ', 'Plu  ']
    start_pos = week.index('Mer  ')
    def is_leap(year): 
            if (not year%4 and year%100 or not year%400): # algorithm to determine leap year
                    return True
    if not is_leap(year):
            del tup[3]
    for month, days in tup: 
        # Print month title
        print('{0} {1}'.format(month, year).center(20, ' '))
                    # Print Day headings
        print(''.join(['{0:<3}'.format(w) for w in week]))
                    # Add spacing for non-zero starting position
        print('{0:<3}'.format('')*start_pos, end='')

        for day in days:
                                # Print day
            print('{0!s:<6}'.format(day), end='')
            start_pos += 1
            if start_pos == 9:
                                        # If start_pos == 9 (Sunday) start new line
                print()
                start_pos = 0 # Reset counter
def make_calendar(year):
    button2 = tk.Button(frame, text="Print Calendar", function=print('
'))
    button2.grid(row=0, column=5)#this makes the buton apear in a grid
start_day=('Mer  ')
make_calendar(year)
root.mainloop()#APPLICATION IS BETWEEN THIS LINE^

Python中的tkinter怎么做？

我如何使用python将终端输出到网格中的tkinter框架？