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How do I check the difference, in seconds, between two dates?

开发者 https://www.devze.com 2023-01-29 04:24 出处:网络
There has to be an easier way to do this. I have objects that want to be refreshed every so often, so I want to record when they were created, check against the current timestamp, and refresh as neces

There has to be an easier way to do this. I have objects that want to be refreshed every so often, so I want to record when they were created, check against the current timestamp, and refresh as necessary.

datetime.datetime开发者_如何学Python has proven to be difficult, and I don't want to dive into the ctime library. Is there anything easier for this sort of thing?


if you want to compute differences between two known dates, use total_seconds like this:

import datetime as dt

a = dt.datetime(2013,12,30,23,59,59)
b = dt.datetime(2013,12,31,23,59,59)

(b-a).total_seconds()

86400.0

#note that seconds doesn't give you what you want:
(b-a).seconds

0


import time  
current = time.time()

...job...
end = time.time()
diff = end - current

would that work for you?


>>> from datetime import datetime

>>>  a = datetime.now()

# wait a bit 
>>> b = datetime.now()

>>> d = b - a # yields a timedelta object
>>> d.seconds
7

(7 will be whatever amount of time you waited a bit above)

I find datetime.datetime to be fairly useful, so if there's a complicated or awkward scenario that you've encountered, please let us know.

EDIT: Thanks to @WoLpH for pointing out that one is not always necessarily looking to refresh so frequently that the datetimes will be close together. By accounting for the days in the delta, you can handle longer timestamp discrepancies:

>>> a = datetime(2010, 12, 5)
>>> b = datetime(2010, 12, 7)
>>> d = b - a
>>> d.seconds
0
>>> d.days
2
>>> d.seconds + d.days * 86400
172800


We have function total_seconds() with Python 2.7 Please see below code for python 2.6

import datetime
import time  

def diffdates(d1, d2):
    #Date format: %Y-%m-%d %H:%M:%S
    return (time.mktime(time.strptime(d2,"%Y-%m-%d %H:%M:%S")) -
               time.mktime(time.strptime(d1, "%Y-%m-%d %H:%M:%S")))

d1 = datetime.now()
d2 = datetime.now() + timedelta(days=1)
diff = diffdates(d1, d2)


Here's the one that is working for me.

from datetime import datetime

date_format = "%H:%M:%S"

# You could also pass datetime.time object in this part and convert it to string.
time_start = str('09:00:00') 
time_end = str('18:00:00')

# Then get the difference here.    
diff = datetime.strptime(time_end, date_format) - datetime.strptime(time_start, date_format)

# Get the time in hours i.e. 9.60, 8.5
result = diff.seconds / 3600;

Hope this helps!


Another approach is to use timestamp values:

end_time.timestamp() - start_time.timestamp()


By reading the source code, I came to a conclusion: the time difference cannot be obtained by .seconds:

@property
def seconds(self):
    """seconds"""
    return self._seconds

# in the `__new__`, you can find the `seconds` is modulo by the total number of seconds in a day
def __new__(cls, days=0, seconds=0, microseconds=0,
            milliseconds=0, minutes=0, hours=0, weeks=0):
    seconds += minutes*60 + hours*3600
    # ...
    if isinstance(microseconds, float):
        microseconds = round(microseconds + usdouble)
        seconds, microseconds = divmod(microseconds, 1000000)
        # ! 
0

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