How to convert these char values to datetime format?
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Hello everyone,
I tried to do this operation to get the duration between both datetimes as follows:
% starting date-time
date_obs = char('2017/09/06');
time_obs = char('11:55:01.109');
% ending date-time
date_end = char('2017/09/06');
time_end = char('12:10:01');
% combining dates & times
datetime_start = strcat(date_obs,{' '},time_obs);
datetime_end = strcat(date_end,{' '},time_end);
%% DATETIMES
datetime1 = datetime(datetime_start,'InputFormat','yyyy/MM/dd HH:mm:ss');
datetime2 = datetime(datetime_end,'InputFormat','yyyy/MM/dd HH:mm:ss');
But I get this error:
Unable to convert '2017/09/06 11:55:01.109' to datetime using the format 'yyyy/MM/dd HH:mm:ss'.
Can you please tell me how to convert them to datetime and get the difference (duration) between both?
I appreciate your help!
Thank you,
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Stephen23
2020년 1월 19일
편집: Stephen23
2020년 1월 19일
The error is caused by the milliseconds in start string: either you need to remove them from the input string, or specify them in the Input format:
>> datetime1 = datetime(datetime_start,'InputFormat','yyyy/MM/dd HH:mm:ss.SSS')
datetime1 =
06-Sep-2017 11:55:01
It is not permitted to have unused characters left over, all characters must correspond to something in the format string. After that it is trivial to calculate the difference (and generate a duration object):
>> dtdiff = datetime2-datetime1
dtdiff =
00:14:59
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Stephen23
2022년 9월 2일
"What will it look like in version 2014a where this feature is missing?"
DATETIME was introduced in R2014b, so you would need to use deprecated date functions, e.g.:
N1 = datenum('2017/09/06 11:55:01.109','yyyy/m/d HH:MM:SS.FFF');
N2 = datenum('2017/09/06 12:10:01','yyyy/m/d HH:MM:SS');
DD = datestr(N2-N1,'HH:MM:SS.FFF')
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