Well, yes, do what the error says -- convert the datetime and/or duration to string representation.
>> dt=datetime(datestr(now)); % create a datetime variable
>> datestr(dt)
dt =
'28-Jun-2022 09:27:41'
>>
The format is/can be set by the 'Format' property of the datetime variable as desired.
>> du=timeofday(datetime(datestr(now))); % for duration
>> string(du)
du =
"09:30:52"
>> char(du)
du =
'09:32:08'
>> cellstr(du)
du =
1×1 cell array
{'09:32:19'}
>> datestr(du)
ans =
' 9:30 AM'
>> timeofday(datetime(datestr(now)))
>>
your choice -- NB: datestr on the duration doesn't preserve the leading zero with the default formatting.
