1

我读了这个并且很感兴趣：Validating date format using regular expression

所以我开始编写自己版本的日期验证功能，我想我很接近，但并不完全，我想要一些建议和提示。我花了很多时间尝试调整功能。

import re
import datetime

# Return True if the date is in the correct format
def checkDateFormat(myString):
    isDate = re.match('[0-1][0-9]\/[0-3][0-9]\/[1-2][0-9]{3}', myString)
    return isDate

# Return True if the date is real date, by real date it means,
# The date can not be 00/00/(greater than today)
# The date has to be real (13/32) is not acceptable
def checkValidDate(myString):
    # Get today's date
    today = datetime.date.today()
    myMaxYear = int(today.strftime('%Y'))

    if (myString[:2] == '00' or myString[3:5] == '00'):
        return False

    # Check if the month is between 1-12
    if (int(myString[:2]) >= 1 or int(myString[:2]) <=12):
        # Check if the day is between 1-31
        if (int(myString[3:5]) >= 1 or int(myString[3:2]) <= 31):
            # Check if the year is between 1900 to current year
            if (int(myString[-4:]) <= myMaxYear):
                return True
    else:
        return False

testString = input('Enter your date of birth in 00/00/0000 format: ')

# Making sure the values are correct
print('Month:', testString[:2])
print('Date:', testString[3:5])
print('Year:', testString[-4:])

if (checkDateFormat(testString)):
    print('Passed the format test')
    if (checkValidDate(testString)):
        print('Passed the value test too.')
    else:
        print('But you failed the value test.')
else:
    print("Failed. Try again")

问题1：int(myString[3:5])当我想比较它是否有效时，还有其他方法（更好）吗？感觉我的方法很重复，而且这个函数必须要00/00/0000，不然会坏掉。因此，从这个意义上说，该功能并不是那么有用。尤其是我处理我的方式00/01/1989，只是简单地比较if他们确实是00。

问题2：有很多if语句，不知道有没有更好的方法来编写这个测试？

我想了解更多关于 python 编程的知识，任何建议或建议将不胜感激。非常感谢。

4

5 回答 5

3

像 python 中的许多东西一样，已经有检查日期的基本功能。假设您不只是将其作为一项学术练习，那么验证日期的最直接方法就是尝试创建它。

import datetime

minyear = 1900
maxyear = datetime.date.today().year

mydate = '12/12/2000'
dateparts = mydate.split('/')
try:
    if len(dateparts) != 3:
       raise ValueError("Invalid date format")
    if int(dateparts[2]) > maxyear or int(dateparts[2]) < minyear:
       raise ValueError("Year out of range")
    dateobj = datetime.date(int(dateparts[2]),int(dateparts[1]),int(dateparts[0]))
except:
    // handle errors

如果给 datetime.date 一个无效的日期，它会抱怨，例如：

datetime.date(2000,45,23)

Traceback (most recent call last):
  File "<pyshell#1>", line 1, in <module>
    datetime.date(2000,45,23)
ValueError: month must be in 1..12
于 2012-04-03T23:12:16.417 回答
1 
from dateutil.parser import *
parse("1993-09-01")
datetime.datetime(1993, 9, 1, 0, 0)

如果格式不正确，则会引发 ValueError，您可以使用 try..catch 捕获它们，以免应用程序意外关闭。

于 2014-07-31T16:22:52.927 回答
1

我知道学术练习有一些价值。我不会劝阻他们。

我还认为学习 Python 的很大一部分是发现已经解决了哪些问题。我有这样的想法有几个原因。首先，我认为重新发明轮子是浪费时间。我也有机会学习代码和解决问题的不同方法。最后，我认为一个长期存在的解决方案不太可能天真地对待主题中的怪癖。

考虑到这一点，我推荐 dateutil 库。

于 2013-04-25T13:53:46.190 回答
1

根据上面的一些答案，我编写了一小段代码，它以以下格式验证日期：DD/MM/YYYY。

date = "29/02/2016"

min_year = 1900
max_year = min_year + 200

days_31 = ['01', '03', '05', '07', '08', '10', '12']
days_30 = ['04', '06', '10', '11']
days_28 = ['02']

month_dict = {'01':'January',
      '02':'February',
      '03':'March',
      '04':'April',
      '05':'May',
      '06':"June",
      '07':'July',
      '08':'August',
      '09':'September',
      '10':'October',
      '11':'November',
      '12':'December'}

def validate(day, month, leap_year_or_not):
    if leap_year_or_not:
        max_day = 29
    else:
        max_day=28
    if month in days_28:
        if day > max_day:
            print "Invalid day: %s for month %s" %(day, month_dict[month])
            return False
        else:
            return True
    elif month in days_30:
        if day > 30:
            print "Invalid day: %s for month %s" %(day, month_dict[month])
            return False
        else:
            return True
    elif month in days_31:
        if day <= 31:
            return True
        else:
            print "Invalid day: %s for month %s" %(day, month_dict[month])
            return False
    else:
        print "Invalid month:%s, Invalid day:%s" %(month, day)
        return False

if len(date)!= 10:
    print "Invalid Format. Please enter date in DD/MM/YYYY"
else:
    day, month, year = date.split("/")
    if len(day) != 2:
        print "Length of day in not 2. Please enter day as 01 for first!" 
    elif len(month) != 2:
        print "Length of month in not 2. Please enter month as 01 for January!"
    elif len(year) != 4:
        print "Length of year in not 4. Please enter year as 2001!"
    else:
        day=int(day)
        month=int(month)
        year=int(year)
        if day < 1 or day > 31:
            print "Day %s is not in range [1-31]" %str(day)
        else:
            if month < 1 or month > 12:
                print "Month %s is not in range [1-12]" %str(month)
            else:
                if year < min_year or year > max_year:
                    print "Year is not in range [%s-%s]" (str(min_year), str(max_year))
            else:
                if year%4 == 0:
                    leap_year = True
                else:
                    leap_year = False
                valid_day_month = validate(day, str(month).zfill(2), leap_year)
于 2016-11-02T15:38:35.157 回答
-1 
import time
import datetime
self.date = raw_input('Enter the date to travel in (yyyy-mm-dd) format: ')
try:
    valid_date = time.strptime(self.date, '%Y-%m-%d')
    today_date=str(datetime.date.today())
    if self.date<today_date:
        print "date got over"
    else:
        print " "
except ValueError:
print('Invalid date!')
于 2015-01-09T03:11:56.823 回答