datetime.timedelta对象代表两个时间之间的的时间差,两个date或datetime对象相减时可以返回一个timedelta对象。
构造函数:
class datetime.timedelta([days[, seconds[, microseconds[, milliseconds[, minutes[, hours[, weeks]]]]]]])
所有参数可选,且默认都是0,参数的值可以是整数,浮点数,正数或负数。
内部只存储days,seconds,microseconds,其他参数的值会自动按如下规则抓转换:
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1 millisecond(毫秒) 转换成 1000 microseconds(微秒)
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1 minute 转换成 60 seconds
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1 hour 转换成 3600 seconds
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1 week转换成 7 days
三个参数的取值范围分别为:
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0 <= microseconds < 1000000
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0 <= seconds < 3600*24 (the number of seconds in one day)
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-999999999 <= days <= 999999999
如果任意参数是float,且小数点后含有microseconds部分,那么microseconds的值为所有参数的微秒部分的总和(四舍五入)如:
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>>> datetime.timedelta(hours=1.232,seconds=20).microseconds
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200000
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>>> datetime.timedelta(hours=1.232,seconds=20.3).microseconds
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500000
支持的操作有:
1 = t2 + t3
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两个timedelta对象相加,同时满足 t1-t2 == t3 and t1-t3 == t2 为True
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t1 = t2 - t3
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两个timedelta对象相减, 同时满足 t1 == t2 - t3 and t2 == t1 + t3 为True
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t1 = t2 * i or t1 = i * t2
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timedelta对象分别乘以i 同时满足 t1 // i == t2 为True, 且 i != 0
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t1 = t2 // i
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向下取整,余数部分被丢弃
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+t1
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返回和t1相同值的timedelta对象
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-t1
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取反操作,等价于(-t1.days, -t1.seconds, -t1.microseconds)和 t1* -1
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abs(t)
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绝对值,等价于: +t 当 t.days >= 0, -t 当 t.days < 0
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str(t)
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返回字符串,格式为: [D day[s], ][H]H:MM:SS[.UUUUUU]
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repr(t)
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返回字符串,格式为: datetime.timedelta(D[, S[, U]])
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此外,timedelta和可以和date,datetime对象进行加减操作,如:
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>>> datetime.datetime.now()
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datetime.datetime(2013, 5, 23, 10, 49, 27, 182057)
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>>> datetime.datetime.now()+datetime.timedelta(2)
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datetime.datetime(2013, 5, 25, 10, 49, 29, 385559)
Python2.7新增了方法:
timedelta.total_seconds()
用于计算秒数。等价于:(td.microseconds + (td.seconds + td.days * 24 * 3600) * 10**6) / 10**6
实例1:
'''时间d距离now()的长度,比如:1分钟前,1小时前,1月前,1年前'''
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import datetime
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def timebefore(d):
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chunks = (
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(60 * 60 * 24 * 365, u'年'),
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(60 * 60 * 24 * 30, u'月'),
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(60 * 60 * 24 * 7, u'周'),
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(60 * 60 * 24, u'天'),
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(60 * 60, u'小时'),
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(60, u'分钟'),
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)
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if not isinstance(d, datetime.datetime):
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d = datetime.datetime(d.year,d.month,d.day)
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now = datetime.datetime.now()
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delta = now - d
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before = delta.days * 24 * 60 * 60 + delta.seconds
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if before <= 60:
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return u'刚刚'
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for seconds,unit in chunks:
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count = before // seconds
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if count != 0:
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break
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return unicode(count)+unit+u"前"
实例2:
‘’‘当前的时间上加一天或一年减一天等操作’‘’
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from datetime import datetime,timedelta
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now = datetime.now()
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yestoday = now - timedelta(days=1)
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tommorow = now + timedelta(days=1)
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next_year = now + timedelta(days = 365)
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