关于实现今天到一年中任意一天两者之间的天数差的计算

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昨天的某时某刻突发奇想，想用自己现阶段所学的python知识来制作一个小程序。大致功能为：开机时像某安全软件一样，弹出窗口，窗口上能提示你，“距xxxx年xx月xx日还有多少天”。

在开始前，因为涉及到日期的计算（如标题所说），首先我想到了标准库中的datetime模块，该模块有个today（）函数能返回今天的日期（格式：xxxx-xx-xx），将函数返回的日期转换成字符串后，日期就成了一个长度为10的字符串，对字符串进行切片后分别转换为整型数，就可得到year，month，day参数，具体代码如下

 1 from datetime import date
 2 
 3 
 4 def get_date():
 5     today_date = date.today()
 6     str_t_d = str(today_date)
 7     year = int(str_t_d[:4])
 8     month = int(str_t_d[5:7])
 9     day = int(str_t_d[-2:])
10     return year, month, day

调用get_date（）后会返回元组（year， month， day）

下一步开始计算天数差了，因为考虑到对于每一年来说，每个月份天数除了2月以外都是固定的，2月天数的变化关系受到年份影响，因此需要判断闰年或平年，代码如下

 1 def is29days(year=get_date()[0]):
 2     months = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
 3     if year % 100 == 0:
 4         if year % 400 == 0:
 5             months[1] = 29
 6             return months
 7         else:
 8             return months
 9     elif year % 4 == 0:
10         months[1] = 29
11         return months
12     else:
13         return months

在上述代码中，将一年的十二个月对应天数按顺序放在列表months中，这个列表中二月对应的天数为28天，即默认为平年，用 if-elif-else语句判断是否为闰年，是就改变months[1]的值为29，返回新的months列表，否返回默认列表，在调用is29days（）时将返回月份列表

有了月份列表，就可也计算了，假定，给定一个日期：3月6日，今天的日期：get_date()[1] 月get_date()[2](今天日期为2月12日，没错我是今天写的)日，利用两个日期的月份对months列表切片，结合列表索引值将其与月份对应，如果当月为2月，则对应的months列表索引值为2-1=1!!!，切片为：months[2-1 ： 3]，这样列表中就包含了2月，3月两个月，用sum计算切片后的列表元素和后，减去两头的天数后就可以得到3月6日与2月12日之间相差的天数了，代码如下

 1 def count_days(num_month=3, d_day=6):
 2     if num_month < get_date()[1] and d_day < get_date()[2]:
 3         '''在上一月的情况'''
 4         msg = 'error'
 5         month = 'error'
 6         year = 'error'
 7         x_day = 'error'
 8         return msg, month, year, x_day
 9     elif num_month == get_date()[1] and d_day < get_date()[2]:
10         '''在当月前些天的情况'''
11         msg = 'error'
12         month = 'error'
13         year = 'error'
14         x_day = 'error'
15         return msg, month, year, x_day
16     else:
17         months = is29days()[get_date()[1]-1:num_month]
18         x_day = d_day
19         year = get_date()[0]
20         month = num_month
21         days = sum(months) - get_date()[2] - months[-1] + x_day
22         msg = '%3s天' % days
23         return msg, month, year, x_day

考虑到会有选择的日期在当日日期之前的情况，加入判断语句，调用count_days（）后返回元组（msg, month, year, x_day），（若不传递实参，则默认实参为num_month=3, d_day=6）

现在就完成了天数差的计算，但开头所讲我们还需一个能弹出窗口（GUI界面），鉴于我仅仅接触过 tkinter 所以我使用了它，代码如下

 1 deadline = Tk()
 2 deadline.title('DeadLine')
 4 deadline.geometry('220x300+1678+732')
 6 label = Label(deadline, text='距%4s年%2s月%2s日还有' % (count_days()[2], count_days()[1], count_days()[3])
 7 label.grid(row=0, columnspan=2)
 8 label_1 = Label(deadline, justify='left', text="""
 9 Don`t waste time any more !
10
11 " So do you want to take a
12 leap of faith，or become an
13 old man，filled with regret
14 waiting to die alone? "
15 """)
16 label_1.grid(row=2, columnspan=2)
17 display = Listbox(deadline, font=100, width=5, height=1)
18 display.insert(END, count_days()[0])
19 display.grid(row=1, columnspan=2)
22 button_2 = Button(deadline, text='Exit', command=deadline.destroy)
23 button_2.grid(columnspan=2)
24 deadline.mainloop()

效果如下

好像差不多了啊

然后开始将其封装成一个.exe的文件，在这里使用第三方库PyInstaller，因为不是python自带的需要自己安装，安装完成后，在控制台输入命令pyinstaller -F -w filename.py，这里需要注意如果不是在filename.py文件所在的目录下打开的控制台，是需要给定路径的，同时F 大写，w小写，如果不加上-w选项，在运行程序时会弹出控制器界面，同时py文件名不能用汉字，否则会报错，完成后在当前目录下，会有一个dist文件件，打开会发现一个filename.exe文件，大功告成!!

还没完，我想每天打开电脑都能看到上图的GUI界面，就像某安全软件在屏幕右下角弹出界面一样（好像安全软件都弹），上网查了一下，找到一个方法，如下

https://jingyan.baidu.com/article/54b6b9c0d103662d593b474e.html

完整的代码如下：

 1 #!/usr/bin/env python3
 2 # coding=utf-8
 3 # Date: 2019/2/12
 4 
 5 from tkinter import *
 6 from datetime import date
 7 
 8 
 9 def get_date():
10     today_date = date.today()
11     str_t_d = str(today_date)
12     year = int(str_t_d[:4])
13     month = int(str_t_d[5:7])
14     day = int(str_t_d[-2:])
15     return year, month, day
16 
17 
18 def is29days(year=get_date()[0]):
19     months = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
20     if year % 100 == 0:
21         if year % 400 == 0:
22             months[1] = 29
23             return months
24         else:
25             return months
26     elif year % 4 == 0:
27         months[1] = 29
28         return months
29     else:
30         return months
31 
32 
33 def count_days(num_month=3, d_day=6):
34     if num_month < get_date()[1] and d_day < get_date()[2]:
35         '''在上一月的情况'''
36         msg = 'error'
37         month = 'error'
38         year = 'error'
39         x_day = 'error'
40         return msg, month, year, x_day
41     elif num_month == get_date()[1] and d_day < get_date()[2]:
42         '''在当月前些天的情况'''
43         msg = 'error'
44         month = 'error'
45         year = 'error'
46         x_day = 'error'
47         return msg, month, year, x_day
48     else:
49         months = is29days()[get_date()[1]-1:num_month]
50         x_day = d_day
51         year = get_date()[0]
52         month = num_month
53         days = sum(months) - get_date()[2] - months[-1] + x_day
54         msg = '%3s天' % days
55         return msg, month, year, x_day
56 
57 
58 deadline = Tk()
59 deadline.title('DeadLine')
60 deadline.geometry('220x300+1678+732')
61 label = Label(deadline, text=
62 '距%4s年%2s月%2s日还有' % (count_days()[2], count_days()[1], count_days()[3]))
63 label.grid(row=0, columnspan=2)
64 label_1 = Label(deadline, justify='left', text="""
65 Don`t waste time any more !
66 
67 " So do you want to take a
68 leap of faith，or become an
69 old man，filled with regret
70 waiting to die alone? "
71 """)
72 label_1.grid(row=2, columnspan=2)
73 display = Listbox(deadline, font=100, width=5, height=1)
74 display.insert(END, count_days()[0])
75 display.grid(row=1, columnspan=2)
76 button_2 = Button(deadline, text='Exit', command=deadline.destroy)
77 button_2.grid(columnspan=2)
78 deadline.mainloop()

写在最后：

　　上述代码只是实现了最简单的情况，我想以此为基础制作一个能计算任意两个日期天数差的程序，一步一步来,day day up，Come on ,code newbie!

posted on 2019-02-12 13:06 大王从来不巡山阅读(...) 评论(...) 编辑收藏