Question

我遇到了python中的日期时间问题。这是我的堆栈跟踪：

28/09/12
Traceback (most recent call last):
  File "product-release-by-year.py", line 49, in <module>
    get_earliest_order_date(fin)
  File "product-release-by-year.py", line 25, in get_earliest_order_date
    date = datetime.datetime.strptime(order_date, '%d/%m/%Y').date()
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/_strptime.py", line 325, in _strptime
    (data_string, format))
ValueError: time data '28/09/12' does not match format '%d/%m/%Y'

相关的代码：

def get_earliest_order_date(fin):
    # CSV Headers
    product_col = 1
    order_col = 0

    f = open(fin, 'rb')
    try:
        reader = csv.reader(f, delimiter=";")
        next(reader, None) 
        for row in reader:

            product_name = row[product_col]

            order_date = row[order_col]
            print order_date

            date = datetime.datetime.strptime(order_date, '%d/%m/%Y').date()

            if product_name not in products:
                products[product_name] = date
            else:
                if date < products[product_name]:
                    products[product_name] = date
    finally:
        f.close()

据我所知，格式字符串应该是正确的？当通过终端直接输入字符串时，对strptime的同一调用也有效。

有什么想法吗？

Answer 1

%Y预计一年有四个数字，包括世纪。使用%y来解析年份为2位的日期。

演示：

>>> import datetime
>>> order_date = '28/09/12'
>>> datetime.datetime.strptime(order_date, '%d/%m/%Y')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/Users/mj/Development/Library/buildout.python/parts/opt/lib/python2.7/_strptime.py", line 325, in _strptime
    (data_string, format))
ValueError: time data '28/09/12' does not match format '%d/%m/%Y'
>>> datetime.datetime.strptime(order_date, '%d/%m/%y')
datetime.datetime(2012, 9, 28, 0, 0)

%y在POSIX convention for C strptime之后增加了一个世纪：

%y
世纪之内（0-99）。什么时候不是一个世纪指定的，69-99范围内的值是指二十年中的年份世纪（1969-1999）; 00-68范围内的值是指二十一世纪（2000-2068）。

Python Strptime格式化

1 个答案: