我想比较两次,如果新的时间超过2分钟,然后if语句将打印输出,我可以得到datetime.datetime.now()的输出,但怎么做我检查旧时间是否少于2分钟?蟒蛇减去时间并运行循环
#!/usr/bin/env python
import datetime
from time import sleep
now = datetime.datetime.now()
sleep(2)
late = datetime.datetime.now()
constant = 2
diff = late-now
if diff <= constant:
print "True time is less than 2min"
else:
print "Time exceeds 2 mins"
有什么想法?
更新:
我现在保存旧日期在文件字符串,然后从当前时间减去它,旧的日期存储格式
2011-12-16 16:14: 50.800856
所以当我做
now = "2011-12-16 16:14:50.838638"
sleep(2)
nnow = datetime.strptime(now, '%Y-%m-%d %H:%M:%S')
late = datetime.now()
diff = late-nnow
它给了我这个错误
ValueError: unconverted data remains: .838638
差异是datetime.timedelta对象,它有.totalseconds()方法或.seconds财产 – yosukesabai 2011-12-16 15:00:43
感谢yosukesabai – krisdigitx 2011-12-16 15:05:43