从python中的日期时间减去秒数
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Subtract seconds from datetime in python
提问by rodrigocf
I have an int variable that are actually seconds (lets call that amount of seconds X). I need to get as result current date and time (in datetime format) minus Xseconds.
我有一个实际上是秒的 int 变量（让我们称之为秒数X）。我需要得到当前日期和时间（以日期时间格式）减去X秒的结果。
Example
例子
If Xis 65 and current date is 2014-06-03 15:45:00, then I need to get the result 2014-06-03 15:43:45.
如果X是 65 并且当前日期是2014-06-03 15:45:00，那么我需要得到结果2014-06-03 15:43:45。
Environment
环境
I'm doing this on Python 3.3.3 and I know I could probably use the datetimemodule but I haven't had any success so far.
我正在 Python 3.3.3 上执行此操作，我知道我可能可以使用该datetime模块，但到目前为止我还没有取得任何成功。
采纳答案by julienc
Using the datetimemodule indeed:
datetime确实使用该模块：
import datetime
X = 65
result = datetime.datetime.now() - datetime.timedelta(seconds=X)
You should read the documentationof this package to learn how to use it!
你应该阅读这个包的文档来学习如何使用它！
回答by wim
>>> from datetime import datetime, timedelta
>>> now = datetime.now()
>>> now
datetime.datetime(2014, 6, 3, 22, 55, 9, 680637)
>>> now - timedelta(seconds=15)
datetime.datetime(2014, 6, 3, 22, 54, 54, 680637)
That's a core library solution, generally for this task I prefer to use a 3rd party class with dateutil.relativedelta, it has more parameters and greater flexibility.
这是一个核心库解决方案，通常对于这个任务，我更喜欢使用 3rd 方类dateutil.relativedelta，它有更多的参数和更大的灵活性。
回答by AlienFromCA
To expand on @julienc's answer, (in case it is helpful to someone)
扩展@julienc的答案，（以防它对某人有帮助）
If you allow X to accept positive or negatives, and, change the subtraction statement to an addition statement, then you can have a more intuitive (so you don't have to add negatives to negatives to get positives) time adjusting feature like so:
如果您允许 X 接受正数或负数，并且将减法语句更改为加法语句，那么您可以拥有更直观的（因此您不必将负数添加到负数以获得正数）时间调整功能，如下所示：
def adjustTimeBySeconds(time, delta):
return time + datetime.timedelta(seconds=delta)
time = datetime.datetime.now()
X = -65
print(adjustTimeBySeconds(time, X))
X = 65
print(adjustTimeBySeconds(time, X))
