Question

date = raw_input()
while date!="END" or date!="end":
    day  = date[0:2]
    month = date[3:5]
    monthsingle = date[3:5]
    monthsingle =str(int(monthsingle))
    monthsingle = int(monthsingle)

我在这里要完成的是将月份编号分配给monthsingle，以便稍后在我的代码中使用它。问题是用户可以输入＆＃34; 02＆＃34;二月。如果没有此错误，我该怎么做：

monthsingle =str(int(monthsingle))
ValueError: invalid literal for int() with base 10: ''

Answer 1

如果用户输入的长度少于四个字符，则date[3:5]将为空字符串。

在尝试将输入字符串转换为整数之前，您可以检查输入字符串是否有效，或者捕获异常并为用户提供有用的错误消息。意外的用户输入不应导致程序崩溃。

while True:
    print('Please enter a date in format "dd/mm" or "end".')
    date = raw_input()  # use input() if you use python 3
    if date.lower() == 'end':
        print('good bye')
        break
    try:
        day = int(date[0:2])
        month = int(date[3:5])
        print('Day is %d and month is %d' % (day, month))
        # Day and month are integers. 
        # You should check that it's a real date as well.
    except ValueError:
        # could not convert to integer
        print('invalid input!')

Answer 2

在此实施中，您可以修剪＆＃39; 0＆＃39; 0在开始时：

monthsingle = date[3:5].lstrip('0')

或查看日期[3：5]：好像有一个点。

Answer 3

It is not a problem that the user is allowed to type 02 for February.错误消息表明问题是monthsingle为空（''）。如果你将一个字符串切成了它的末尾;你得到一个空字符串。这意味着输入不是dd/mm格式。

要在不使用datetime.strptime()功能的情况下解析日期：

while True:
    try:
        day, month = map(int, raw_input("Enter date dd/mm: ").split('/'))
        # validate day, month here...
    except ValueError:
        print 'invalid input, try again'
    else:
        break

# use day, month to get the next date...
# you could use datetime module to check your answer:
from datetime import date, timedelta
print(date(date.today().year, month, day) + timedelta(1))

Answer 4

def generate_next_date(day,month,year):
    #Start writing your code here
    if((year%400==0 or year%4==0) and month==2):
        next_day=day+1
        next_month=month
        next_year=year
    elif(month==2 and day==28):
        next_day=1
        next_month=month+1
        next_year=year
    elif(month==12 and day==31):
        next_day=1
        next_month=1
        next_year=year+1
    elif(day==31 ):
        next_day=1
        next_month=month+1
        next_year=year
    elif((day==30) and (month==4 or month==6 or month==9 or month==11)):
        next_day=1
        next_month=month+1
        next_year=year
    else:
        next_day=day+1
        next_month=month
        next_year=year




    print(next_day,"-",next_month,"-",next_year)


generate_next_date(28,2,2015)

Answer 5

Java 中的解决方案

<块引用>

实现一个程序来生成和显示给定日期的下一个日期。日期将以日、月和年的形式提供，如下表所示。输出应以以下格式显示：日-月-年。假设：输入将始终是一个有效的日期。

class NextDate 
{
    public static void main(String[] args) 
    {
        // Implement your code here 
        int day = 31,month = 12, year=15,monthLength;
        year = 2000+year;
        char leap;
        
        if((year % 4==0) & (year % 100 == 0) || (year % 400 == 0))
        {
            leap = 'y';
        }
        else
        {
            leap = 'n';
        }

        if(month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12)
        {
            monthLength = 31;
        }
        else if(month == 2)
        {
            if(leap == 'y')
            {
                monthLength = 28;
            }
            else
            {
                monthLength = 29;
            }
            
        }
        else
        {
            monthLength = 30;
        }
        
        if(day < monthLength)
        {
            day++;
        }
        else if(month == 12 & day == monthLength)
        {
            day = 1;
            month=1;
            year++;
        }
        System.out.println(day+"-"+month+"-"+year);
    }
}

给定日期，要求打印下一个日期

5 个答案: