Python中对时间日期的处理方法简单汇总

这篇文章主要介绍了Python实用日期时间处理方法汇总,本文讲解了获取当前datetime、获取当天date、获取明天/前N天、获取当天开始和结束时间(00:00:00 23:59:59)、获取两个datetime的时间差、获取本周/本月/上月最后一天等实用方法 ,需要的朋友可以参考下

原则, 以datetime为中心, 起点或中转, 转化为目标对象, 涵盖了大多数业务场景中需要的日期转换处理

步骤:

1. 掌握几种对象及其关系 2. 了解每类对象的基本操作方法 3. 通过转化关系转化 涉及对象

1. datetime

代码如下: >>> import datetime >>> now = datetime.datetime.now() >>> now datetime.datetime(2015, 1, 12, 23, 9, 12, 946118) >>> type(now) <type 'datetime.datetime'>

2. timestamp

代码如下: >>> import time >>> time.time() 1421075455.568243

3. time tuple

代码如下: >>> import time >>> time.localtime() time.struct_time(tm_year=2015, tm_mon=1, tm_mday=12, tm_hour=23, tm_min=10, tm_sec=30, tm_wday=0, tm_yday=12, tm_isdst=0)

4. string

代码如下: >>> import datetime >>> datetime.datetime.now().strftime("%Y-%m-%d %H:%M:%S") '2015-01-12 23:13:08'

5. date

代码如下: >>> import datetime >>> datetime.datetime.now().date() datetime.date(2015, 1, 12)

datetime基本操作

1. 获取当前datetime

代码如下: >>> import datetime >>> datetime.datetime.now() datetime.datetime(2015, 1, 12, 23, 26, 24, 475680)

2. 获取当天date

代码如下: >>> datetime.date.today() datetime.date(2015, 1, 12)

3. 获取明天/前N天

明天

代码如下: >>> datetime.date.today() + datetime.timedelta(days=1) datetime.date(2015, 1, 13)

三天前

代码如下: >>> datetime.datetime.now() datetime.datetime(2015, 1, 12, 23, 38, 55, 492226) >>> datetime.datetime.now() - datetime.timedelta(days=3) datetime.datetime(2015, 1, 9, 23, 38, 57, 59363)

4. 获取当天开始和结束时间(00:00:00 23:59:59)

代码如下: >>> datetime.datetime.combine(datetime.date.today(), datetime.time.min) datetime.datetime(2015, 1, 12, 0, 0) >>> datetime.datetime.combine(datetime.date.today(), datetime.time.max) datetime.datetime(2015, 1, 12, 23, 59, 59, 999999)

5. 获取两个datetime的时间差

代码如下: >>> (datetime.datetime(2015,1,13,12,0,0) - datetime.datetime.now()).total_seconds() 44747.768075

6. 获取本周/本月/上月最后一天

本周

代码如下: >>> today = datetime.date.today() >>> today datetime.date(2015, 1, 12) >>> sunday = today + datetime.timedelta(6 - today.weekday()) >>> sunday datetime.date(2015, 1, 18)

本月

复制代码 代码如下: >>> import calendar >>> today = datetime.date.today() >>> _, last_day_num = calendar.monthrange(today.year, today.month) >>> last_day = datetime.date(today.year, today.month, last_day_num) >>> last_day datetime.date(2015, 1, 31)

获取上个月的最后一天(可能跨年)

代码如下: >>> import datetime >>> today = datetime.date.today() >>> first = datetime.date(day=1, month=today.month, year=today.year) >>> lastMonth = first - datetime.timedelta(days=1)

关系转换

几个关系之间的转化

Datetime Object / String / timestamp / time tuple

关系转换例子

datetime <=> string

datetime -> string

代码如下: >>> import datetime >>> datetime.datetime.now().strftime("%Y-%m-%d %H:%M:%S") '2015-01-12 23:13:08'

string -> datetime

代码如下: >>> import datetime >>> datetime.datetime.strptime("2014-12-31 18:20:10", "%Y-%m-%d %H:%M:%S") datetime.datetime(2014, 12, 31, 18, 20, 10)

datetime <=> timetuple

datetime -> timetuple

代码如下: >>> import datetime >>> datetime.datetime.now().timetuple() time.struct_time(tm_year=2015, tm_mon=1, tm_mday=12, tm_hour=23, tm_min=17, tm_sec=59, tm_wday=0, tm_yday=12, tm_isdst=-1)

timetuple -> datetime

代码如下: timetuple => timestamp => datetime [看后面datetime<=>timestamp]

datetime <=> date

datetime -> date

代码如下: >>> import datetime >>> datetime.datetime.now().date() datetime.date(2015, 1, 12)

date -> datetime

代码如下: >>> datetime.date.today() datetime.date(2015, 1, 12) >>> today = datetime.date.today() >>> datetime.datetime.combine(today, datetime.time()) datetime.datetime(2015, 1, 12, 0, 0) >>> datetime.datetime.combine(today, datetime.time.min) datetime.datetime(2015, 1, 12, 0, 0)

datetime <=> timestamp

datetime -> timestamp

代码如下: >>> now = datetime.datetime.now() >>> timestamp = time.mktime(now.timetuple()) >>> timestamp 1421077403.0

timestamp -> datetime

代码如下: >>> datetime.datetime.fromtimestamp(1421077403.0) datetime.datetime(2015, 1, 12, 23, 43, 23)

附加

1. #当月1号   datetime.date(datetime.date.today().year,datetime.date.today().month,1)  
2. #当月1号  datetime.date.today().replace(day=1)  
3. #上月1号  (datetime.date.today().replace(day=1) - datetime.timedelta(1)).replace(day=1)