How to convert these char values to datetime format?

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Mohamed Nedal
Mohamed Nedal on 19 Jan 2020
Commented: Mohamed Nedal on 19 Jan 2020
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);
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,

Accepted Answer

Stephen on 19 Jan 2020
Edited: Stephen on 19 Jan 2020
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 =

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