Whiel Update ,datetime value is rounded to seconds.! i want milliseconds too
Tag : sql , By : WuJanJai
Date : March 29 2020, 07:55 AM
I wish did fix the issue. In the case where you're doing a straight update of a datetime in one table with one from another table (i.e. no fiddling with the value), then it sounds like the datatype in the table being updated is not the same.
i.e. in SQL Server world, it could be that you are using SMALLDATETIME column in the table being updated, but a DATETIME field in the table being copied from. SMALLDATETIME is only accurate to the second and so would show this behaviour
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Remove hours:seconds:milliseconds in DateTime object
Tag : chash , By : Tony Z
Date : March 29 2020, 07:55 AM
hop of those help? To answer your question, no - you would have to store it in a different type. The most simple choice is to use a string. string date = dateTime.ToString("MM:dd:yyyy");
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How to get min, seconds and milliseconds from datetime.now() in python?
Date : March 29 2020, 07:55 AM
this will help , What about: datetime.now().strftime('%M:%S.%f')[:-4] now=datetime.now()
string_i_want=('%02d:%02d.%d'%(now.minute,now.second,now.microsecond))[:-4]
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Is Perl DateTime epoch in milliseconds or seconds?
Date : March 29 2020, 07:55 AM
it fixes the issue In the context of DateTime, "epoch" refers to Unix time, which is the number of seconds that aren't leap seconds since 1970-01-01T00:00:00Z. $ perl -MDateTime -e'CORE::say DateTime->from_epoch( epoch => 1 )'
1970-01-01T00:00:01
$ perl -MDateTime -e'CORE::say DateTime->from_epoch( epoch => 1000 )'
1970-01-01T00:16:40
my $js_time = $dt->epoch * 1000 + $dt->millisecond;
my $dt = DateTime->now();
use Time::HiRes qw( time );
my $dt = DateTime->from_epoch( epoch => time() );
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Datetime milliseconds to seconds in Pandas
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Use floor with T for minutes for set 0 seconds: #if necessary
#df['time'] = pd.to_datetime(df['time'])
df['time'] = df['time'].dt.floor('T')
#alternative solution
#df['time'] = df['time'].dt.floor('Min')
print (df)
time
0 2018-04-11 22:18:00
1 2018-04-11 23:00:00
df['time'] = df['time'].dt.round('T')
print (df)
time
0 2018-04-11 22:19:00
1 2018-04-11 23:00:00
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