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Solved
Check time between two TDateTimes
Posted on 1998-07-19
OK, I've started making a simple alarm program and I've run into a fairly major problem - I'm saving all my alarms OK and successfully recalling them as TDateTimes, but there's a difficulty. Are you sitting comfortably?
Say I have alarms at 08:30, 11:15 and 16:30. This is fine if the app starts before the earliest alarm - it can progress through the sequence easily enough. But since the app runs on a network, it is getting started and stopped all the time, so if a user logs on and the app runs at 10:13, it needs to reorganize the order in which the alarms come.
Bear in mind that this has to be flexible: a user might stay logged on for over 24 hours...
I've tried doing stuff like Now-AlarmOne but mathematical operators seem to go haywire with this!
Good Luck!
Ed
Say I have alarms at 08:30, 11:15 and 16:30. This is fine if the app starts before the earliest alarm - it can progress through the sequence easily enough. But since the app runs on a network, it is getting started and stopped all the time, so if a user logs on and the app runs at 10:13, it needs to reorganize the order in which the alarms come.
Bear in mind that this has to be flexible: a user might stay logged on for over 24 hours...
I've tried doing stuff like Now-AlarmOne but mathematical operators seem to go haywire with this!
Good Luck!
Ed
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6 Comments
The TDateTime format allows you to compare times. You can write a thing like this: if time1> time2 then... Do you understand? It will compare the first time and the seccond.
Can you exaplin better what is your problem and what you don't understand?
Can you exaplin better what is your problem and what you don't understand?
Most mathematical operators should work fine with TDateTime.
The most commonly used is -
But you can also / * and + with it.
(Time2-Time1)*24*60 is the amount of minutes passed between Time1 and Time2.
A TDateTime is just a Double.
The whole part are the days since 1899.
The fraction is the part of the day.
1.5 =
1 day
12 hours
1 hour is 1/24
1 minute 1/(24*60)
1 second 1/(24*60*60)
You can also use DecodeDate EncodeDate DecodeTime and EncodeTime.
Regards Jacco
The most commonly used is -
But you can also / * and + with it.
(Time2-Time1)*24*60 is the amount of minutes passed between Time1 and Time2.
A TDateTime is just a Double.
The whole part are the days since 1899.
The fraction is the part of the day.
1.5 =
1 day
12 hours
1 hour is 1/24
1 minute 1/(24*60)
1 second 1/(24*60*60)
You can also use DecodeDate EncodeDate DecodeTime and EncodeTime.
Regards Jacco
Could you define your problem more?
Lets say a user logs on at 10:13, he had an alarm set at 08:30 do you wont that alarm to go off?
Should the 08:30 alarm go off every day or only on the set date?
Let me know and I will help you.
Lets say a user logs on at 10:13, he had an alarm set at 08:30 do you wont that alarm to go off?
Should the 08:30 alarm go off every day or only on the set date?
Let me know and I will help you.
? Time and date are stored in a float type (or is it double?). Heck doesn't mind, either way the date is defined with the numbers before the decimal separator and time is defined with these after the decimal sperator. Now when you compare two times it is fairly simple isn't it? the number later on the day will always be greater then the other....
Or am I missing the question here?
Zif.
Or am I missing the question here?
Zif.
ok, several people so far have said that the date is in the integer part and the time in the frac-part.
The reason I'm posting this as an answer is because I think I see your problem: These are DAILY alarms, right? 8:30 every day?
You can "solve" your problems by using the frac-part of your times, thereby ignoring the date/day.
IF (Frac(Now)>Frac(Alarm)) // later than alarm
AND (Frac(Now)<Frac(Alarm)+5/2
THEN SoundAlarm;
// btw: "now" is a function - instead of calling it several times, store result in a variable. Also reduces strange effects if "Now" changes from line to line *g* - it rarely hits a critical value, but if it only happens once in a thousand times, it's almost impossible to track down :-))
Of course you have to "remember"/store which alarms were used and which ones weren't...
I don't really see a need to "sort" the alarms, but if you really want to:
SecsRemaining:=Round( (Frac(Alarm)-Frac(Now)+1)*
// if you are using 16-bit Delphi, you might need to typecast to avoid overflows
// this should return the seconds remaining before the alarm should be triggered the next time. Use this value to sort.
Good luck! Let me know if that helped and ask for details before rejecting the answer.
:o))
It's ( '-)
:-))
The reason I'm posting this as an answer is because I think I see your problem: These are DAILY alarms, right? 8:30 every day?
You can "solve" your problems by using the frac-part of your times, thereby ignoring the date/day.
IF (Frac(Now)>Frac(Alarm)) // later than alarm
AND (Frac(Now)<Frac(Alarm)+5/2
THEN SoundAlarm;
// btw: "now" is a function - instead of calling it several times, store result in a variable. Also reduces strange effects if "Now" changes from line to line *g* - it rarely hits a critical value, but if it only happens once in a thousand times, it's almost impossible to track down :-))
Of course you have to "remember"/store which alarms were used and which ones weren't...
I don't really see a need to "sort" the alarms, but if you really want to:
SecsRemaining:=Round( (Frac(Alarm)-Frac(Now)+1)*
// if you are using 16-bit Delphi, you might need to typecast to avoid overflows
// this should return the seconds remaining before the alarm should be triggered the next time. Use this value to sort.
Good luck! Let me know if that helped and ask for details before rejecting the answer.
:o))
It's ( '-)
:-))
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