Question

在我的申请中,我正在获得时间投入,其形式是:Values = 12,12,12。

Now i need to validate it h<24 M<60 S<60 etc. and i want final output in %H:%M:%S format.
To get this i tried datetime.time().

我曾用<代码>价值 = 12 然后用<代码>12,12,尝试过。

In [1]: import datetime

In [2]: values =  12 

In [3]: d = datetime.time(values)

TypeError  Traceback (most recent call last)
/mypc/test/<ipython console> in <module>()
TypeError: an integer is required

In [4]: d = datetime.time(int(values))

In [5]: d
Out[5]: datetime.time(12, 0)

In [6]: d.strftime( %H:%M:%S )
Out[6]:  12:00:00 

In [7]: s = d.strftime( %H:%M:%S )

In [8]: s
Out[8]:  12:00:00 

In [9]: values =  12,12,12 

In [10]: d = datetime.time(int(values))

ValueError: invalid literal for int() with base 10:  12,12,12

但它的工作如下。

In [24]: datetime.time(12,12,12).strftime( %H:%M:%S )
Out[24]:  12:12:12

因此,问题是这一时间。时间()作为星体和<代码>12,12,的输入不能按字母换算。

Is there any other way(otherthan regexp) to do the validation for only Hr:M:S.

Answer 1

You have to unpack the values:

>>> values =  12,12,12 
>>> values = ast.literal_eval(values)
>>> datetime.time(*values)
datetime.time(12, 12, 12)

如果给付的时间无效,最后一份声明将产生错误。

为了避免出现零增量的问题,正如“wim”所指出的,有可能将第二行改为:

values = (int(i) f或i in values.split( , ))

或

values = map(int, values.split( , ))

Answer 2

请注意,<代码>strftime的对口单位为strptime,即,如果你需要一段时间才能使用strptime。采用与你使用<代码>strftime相同的格式,用该格式印刷一段时间:

>>> import time
>>> t = time.strptime( 12,12,12 ,  %H,%M,%S )
>>> time.strftime( %H:%M:%S , t)
 12:12:12

这样,strptime (H<24, M<60, S<=61):

>>> time.strptime( 24,0,0 ,  %H,%M,%S )
...
ValueError: time data  24,0,0  does not match format  %H,%M,%S 
>>> time.strptime( 0,60,0 ,  %H,%M,%S )
...
ValueError: time data  0,60,0  does not match format  %H,%M,%S 
>>> time.strptime( 0,0,62 ,  %H,%M,%S )
...
ValueError: unconverted data remains: 2

Note that strptime allows S<=61 as explained in the documentation:

The range really is 0 to 61; this accounts for leap seconds and the (very rare) double leap seconds.

如果这给你带来问题,那么你很可能需要将这一价值纳入你的法典。

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