python - 在Tkinter中尝试约会-Python

Question

经过多次测试并阅读了数小时的Stackoverflow之后，我决定纠正这个问题。我的文字(较大代码的一部分)如下:

import pandas as pd
import datetime as dt
from tkinter import *
import tkinter.filedialog
from tkinter import messagebox


def test_click():
  global ipt_dt   
  global coef
  global z
  global w
  z = item_prosp['Accrual_Start'].min()
  w = item_prosp['Accrual_End'].max()
  ipt_d = tkvar_d.get()
  ipt_m = tkvar_m.get()
  ipt_y = tkvar_y.get()  
  x = 0
  while x == 0:
    ipt = str(ipt_d + '/'+ ipt_m + '/' + ipt_y)
    try:
        ipt_dt = dt.datetime.strptime(ipt, "%d/%b/%Y")
        if ipt_dt < z or ipt_dt > w:
                messagebox.showinfo("Error", "The input date is outside scope date") 
    else:
         print("Date ok")
        x =+ 1
    except: 
        messagebox.showerror("Error", "The input date is not valid")
        ipt_d = 0
        ipt_m = 0
        ipt_y = 0
        continue

生成输入的代码的tkinter部分为:

 #Question 1 - Evaluation date
 label4 = Label(window, text='Please inform the valuation date :', bg='white').grid(row=13, column=0, columnspan=3, pady=2, sticky=W)
 tkvar_d = StringVar(window)
 tkvar_m = StringVar(window)
 tkvar_y = StringVar(window)
 choices_d = ['1', '2', '3','4','5','6','7','8','9','10','11','12','13','14','15','16','17','18','19','20','21','22','23','24','25','26','27','28','29','30','31']
 choices_m = ['Jan', 'Feb', 'Mar','Apr','May','Jun','Jul','Aug','Sep','Oct', 'Nov', 'Dec']
 choices_y = ['2018','2019', '2020', '2021','2022','2023','2024','2025','2026','2027','2028','2029','2030']
 popupmenu_d = OptionMenu(window, tkvar_d, *choices_d)
 popupmenu_m = OptionMenu(window, tkvar_m, *choices_m)
 popupmenu_y = OptionMenu(window, tkvar_y, *choices_y)
 label5 = Label(window, text='Day :', bg='white').grid(row=14, column=0, sticky=E+W)
 popupmenu_d.grid(row=15, column=0, padx=2, sticky=E+W)
 label6 = Label(window, text='Month :', bg='white').grid(row=14, column=1, sticky=E+W)
 popupmenu_m.grid(row=15, column=1, padx=2, sticky=E+W)
 label7 = Label(window, text='Year :', bg='white').grid(row=14, column=2, sticky=E+W)
 popupmenu_y.grid(row=15, column=2, padx=2, sticky=E+W)
 Button(window, text="Test Date", width=10, command=test_click).grid(row=15, column=3, padx=5, pady=10, sticky=W)

运行文件时W的样本值是:

2018-04-18 00:00:00

对于Z是:

2018-04-18 00:00:00

我需要导入一个文件(外部构建并已结构化)，从中读取2个值(代码中的变量Z和W)，并将其与输入变量(ipt_dt)进行比较，该变量是用户通过3填写的日期tkinter的下拉菜单。

错误是try不会通过if语句，并且如果输入不在合并范围日期之内，它将永远不会输出。每当我输入的日期小于最小日期或大于最大日期时，如果将日期设置为“Date ok”，它将返回showerror消息。

任何人都对如何解决这个问题有任何想法，或者为什么我的fi被忽略了？

谢谢!

Answer 1

我查看了您最初发布的代码，然后使用load_click函数将Excel加载到df中。但是实际上您不会在任何地方运行load_click函数，因此不会加载数据帧，因此不会填充z和w。

如果您按如下方式更改click1()函数，则它应该可以工作(它对我来说有一些示例数据)。

def click1():
  global a
  a = tkinter.filedialog.askopenfilename(initialdir = "/",title = "Select file", filetypes = ( ("Excel file", "*.xlsx"), ("All files", "*.*") ) )
  output1.insert(END, a)
  global a1
  a1 = output1.get() 
  load_click()

或根据需要添加一个单独的“加载”按钮(在＃File1部分的底部):

Button(window, text="Load", width=6, command=load_click).grid(row=4, column=3, padx=5, sticky=W)

您可能还想在if语句中添加另一个x = 1。否则，由于while循环，消息框将继续 pop ，从而无法更正输入日期。

x = 0
while x == 0:
    ipt = str(ipt_d + '/'+ ipt_m + '/' + ipt_y)
    try:
        ipt_dt = dt.datetime.strptime(ipt, "%d/%b/%Y")
        print type(ipt_dt)
        if (ipt_dt < z) or (ipt_dt > w):
            messagebox.showinfo("Error", "The input date is outside scope date")
            x = 1 # I've added this one
        else:
            print("Date ok")
            x =+ 1
    except: 
        messagebox.showerror("Error", "The input date is not valid")
        ipt_d = 0
        ipt_m = 0
        ipt_y = 0
    continue