python - 日期时间和 utctimetuple()

Question

在此代码中，3 个项目 a、b、z 的 mDATE0 为:

mUNIQUE: z | mDATE0: 2010-11-14 14:55:04.293000
mUNIQUE: b | mDATE0: 2010-11-14 14:53:34.824000
mUNIQUE: a | mDATE0: 2010-11-14 14:50:14.155000

但是当我这么做的时候

...
utc_tuple = rep.mDATE0.utctimetuple()
...

对应的utc_tuples是:

utc_tuple: time.struct_time(tm_hour=14, tm_min=55)
utc_tuple: time.struct_time(tm_hour=14, tm_min=55)
utc_tuple: time.struct_time(tm_hour=14, tm_min=55)

换句话说，min=55 适用于所有项目，而 mDATE0 具有

z --> min=55; 
b --> min=53; 
a --> min=50

我做错了什么？请看我的related question 。谢谢。

Answer 1

那不是代码!请发布简短、完整的示例代码来说明问题。以下是我认为您正在尝试做的事情，但是如果没有看到您的代码，任何人都无法指出您的错误。

from datetime import datetime
# build up some datetime objects.
z = datetime.strptime('2010-11-14 14:55:04.293000','%Y-%m-%d %H:%M:%S.%f')
b = datetime.strptime('2010-11-14 14:53:34.824000','%Y-%m-%d %H:%M:%S.%f')
a = datetime.strptime('2010-11-14 14:50:14.155000','%Y-%m-%d %H:%M:%S.%f')
# display them
print 'z =',z
print 'b =',b
print 'a =',a
# print the minute
print 'z min =',z.utctimetuple().tm_min
print 'b min =',b.utctimetuple().tm_min
print 'a min =',a.utctimetuple().tm_min
# print the minute an easier way
print 'z min =',z.minute
print 'b min =',b.minute
print 'a min =',a.minute

输出:

z = 2010-11-14 14:55:04.293000
b = 2010-11-14 14:53:34.824000
a = 2010-11-14 14:50:14.155000
z min = 55
b min = 53
a min = 50
z min = 55
b min = 53
a min = 50