Run one function only once when the function starts everyday or some time period
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Hi,
I want to do this: with my login system, I only want it to run once when the user starts to use it everyday or some time let's say 5 hours etc... not everytime when people run the funtion you need to log in every time.
For example this login function is inside one big function, so everytime when you run this big function, it can only run the login once everyday. So even let's say I run the function 3 times within 5 minutes, I do not want to login every time I run. Only the first time is enough....I know a counter might work, like to change from 0 to 1, but then later how can I clear it next day? Or maybe PERSISTENT...anyone can help ?
thanks
2 Comments
Titus Edelhofer
on 3 Feb 2015
I have a rough idea what you want to achieve, but maybe you can give some more details ...? Would make giving a recommendation easier.
Titus
Accepted Answer
Guillaume
on 3 Feb 2015
Edited: Guillaume
on 4 Feb 2015
Yes, use persistent and store the last time it was run in your variable. Then compare the current time to the stored time and if not enough time has elapsed just return from the function:
function login()
persistent rundate;
lastrun = rundate;
rundate = now;
if ~isempy(lastrun) && hours(datetime(rundate, 'ConvertFrom', 'datenum') - datetime(lastrun, 'ConvertFrom', 'datenum')) < 24
return;
end
%... normal content of login
end
7 Comments
Guillaume
on 4 Feb 2015
Edited: Guillaume
on 4 Feb 2015
Yes, it is in days. As I said you need to understand how datenum are constructed. The integral part is the number of days (since a reference time), the fractional is made up of hours/24 + minutes/(24*60) + seconds/(24*60*60) + milliseconds/(24*60*60*60).
The new datetime and duration types of R2014b make it much easier to extract a duration in whichever unit of time you want.
More Answers (1)
Titus Edelhofer
on 3 Feb 2015
Hi,
you mean something like this:
function Data()
l = login();
code of data processing....etc
function aLogin = login()
persistent theLogin
persistent lastLogin
if isempty(theLogin)
% first call
theLogin = doSomething;
lastLogin = now;
elseif (now-lastLogin) > 2 / 24
% if last login is more than two hours ago
theLogin = doSomething;
end
aLogin = theLogin;
end
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