datevec back to regular time
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I am using datevec and performing calulations. The end result needs to be converted back to a regular time for the user. I have tried different things and I am still am getting the result I want. I did not put a date in which I normally do using the Date Picker. Can you help me?
t=datevec(state.TimeHHMMssEditField); % where state.TimeHHMMssEditField = 1:31:00
desiredTime = t(4) *3600 + t(5) * 60 + t(6);
NewTime =desiredTime % 5460
StartTime = NewTime + .5 %5460.5
Duration = (state.DurationinsecondsEditField)' % where state.DurationinsecondsEditField = 30 seconds
EndTime = StartTime + Duration %5490.5
% EndTime is in seconds. I want to convert the EndTime (5490.5) to look like 1:31:35
How do I do this?
采纳的回答
Walter Roberson
2020-11-9
datestr(EndTime/(3600*24),'HH:MM:ss')
However, this will have leading 0 . Leading 0 is used automatically if you do not use AM indicator.
I would recommend, by the way, that you switch to using datetime and duration objects, which make the code a lot cleaner.
5 个评论
Jo Bremer Balestracci
2020-11-9
Jo Bremer Balestracci
2020-11-9
Walter Roberson
2020-11-9
You did not add 5 seconds, you added 0.5 seconds. That 0.5 seconds is not visible because you did not ask to display fractions of a second.
Jo Bremer Balestracci
2020-11-9
Walter Roberson
2020-11-9
'HH:MM:ss.ff' for serial datenum work.
'HH:mm:ss.SS' for datetime() work.
更多回答（1 个）
I second Walter Roberson's suggestion to skip the datevec form and to use duration.
t = '1:31:00'
du = duration(t)
newtime = du + seconds(0.5)
newtime.Format = 'hh:mm:ss.SSS' % Show fractional seconds
newt = string(newtime) % or char(newtime)
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