datevec back to regular time

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?

Answer 1

[Walter Roberson]

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.

[Jo Bremer Balestracci]

This did the format conversion but I did get 1:31:35. I got 1:31:30 so I lost 5 seconds. I am going to give it annother try and see what the outcome is

[Jo Bremer Balestracci]

Why am I missing the .5 that was added to the StartTime variable??

[Walter Roberson]

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]

How do I diplay fractions of a second? Do I use HH:MM:ss:SS?

[Walter Roberson]

'HH:MM:ss.ff' for serial datenum work.

'HH:mm:ss.SS' for datetime() work.

Answer 2

[Steven Lord]

I second Walter Roberson's suggestion to skip the datevec form and to use duration.

t = '1:31:00'

t = '1:31:00'

du = duration(t)

du = duration
   01:31:00

newtime = du + seconds(0.5)

newtime = duration
   01:31:00

newtime.Format = 'hh:mm:ss.SSS' % Show fractional seconds

newtime = duration
   01:31:00.500

newt = string(newtime) % or char(newtime)

newt = "01:31:00.500"