TheGameCreators - CodeBase - Duration calculator

Category: Functions

Version: 2.0

Information

Uploaded: 3rd Feb 2013 22:05

Modified: 12th Jul 2013 21:56

Author: HWT

Summary

- Calculates duration between 2 sets of date-time. - Returns time elapsed in seconds. - Useful for projects involving time calculation based on system time. E.g. games like Animal Forest where activity must occur even after exiting the game.

Full Description

DarkBASIC Professional Source Code

` This code was downloaded from The Game Creators

` It is reproduced here with full permission

` http://www.thegamecreators.com

Rem Project: elapsed time

Rem Created: Monday, January 28, 2013

Rem Author: Damian R Gamlath

// Note: If you find any bugs at all, please let me know by email (helloworldtomorrow@yahoo.com).

// Also, if you find it useful and include it, I would really appreciate being mention in the credits :-)

`INSTRUCTIONS:

`Date format: mm/dd/yyyy e.g. 02/23/1990

`Time format: hh:mm:ss e.g. 14:01:59

`Data must be input in string format e.g. "02/23/1990" and "21:01:57"

`Returns duration elapsed in seconds

`Currently does not support old date < new date i.e. calculating past date MINUS future date.

`example only...

a = _duration("02/03/2013","12/28/2026","23:59:59","15:14:10")

wait key

end

function _duration(old_date$,new_date$,old_time$,new_time$)

`** EXTRACT DATA FROM TEXT **

`Hours

new_hr = val(left$(new_time$,2))

old_hr = val(left$(old_time$,2))

`Minutes

new_min = val(right$(left$(new_time$,5),2))

old_min = val(right$(left$(old_time$,5),2))

`SEconds

new_sec = val(right$(new_time$,2))

old_sec = val(right$(old_time$,2))

`Days

new_day = val(right$((left$(new_date$,5)),2))

old_day = val(right$((left$(old_date$,5)),2))

`Months

new_mon = val(left$(new_date$,2))

old_mon = val(left$(old_date$,2))

`Years

new_yr = val(right$(new_date$,4))

old_yr = val(right$(old_date$,4))

`** TO FIND DAYS **

If old_yr = new_yr

`output

Print "IN THE SAME YEAR.." : print ""

`FIRST: use time differences to find the hours/seconds/minutes between the two times

`NOTE: these commands MUST be in this order - because we are advancing old time as we go along, if you put the hours command first, it will advance minutes BUT!!

`in reality, the MINUTES will impact the HOURS not the other way around - hence we must put seconds -> minutes -> hours -> days -> months -> years

SEC = (60 - old_sec) + new_sec : old_min = old_min + 1

MIN = (60 - old_min) + new_min : old_hr = old_hr + 1

HRS = (24 - old_hr) + new_hr : old_day = old_day + 1

If SEC >= 60 then SEC = SEC - 60 : MIN = MIN + 1

If MIN >= 60 then MIN = MIN - 60 : HRS = HRS + 1

If HRS >= 24 then HRS = HRS - 24 : DDD = DDD + 1

`Wihin the same month and year THEN find DAYS by just adding the difference between new day and old day (we must ADD the difference

`because otherwise it will neglect any values obtained for DDD from the time section above

If old_mon = new_mon then DDD = DDD + (new_day - old_day)

`if the difference in months is just 1 (so the months are right next to each other)

`NOTE i put in > because in the routine below (where i check for the days found in the months BETWEEN dates...

`i need to ALSO find the days remaining in the initial and FINAL dates. instead of repeating this script - i just put in the >= sign

If abs(new_mon - old_mon) >= 1

`... then just find how many days left in the month and add that to the days elapsed in the new month

`you need to add one because of the time section above - it manipulates old_day - ALSO, when you subtract

`28 - 2 for example you will get 26 BUT you need to include 28th (which is not done because of subtraction) so you add 1 to compensate

`for some reason, i must add one to the final DDD value

If old_mon = 1 Then DDD = DDD + (31 - old_day) + new_day

If old_mon = 2 Then DDD = DDD + (28 - old_day) + new_day

If old_mon = 3 Then DDD = DDD + (31 - old_day) + new_day

If old_mon = 4 Then DDD = DDD + (30 - old_day) + new_day

If old_mon = 5 Then DDD = DDD + (31 - old_day) + new_day

If old_mon = 6 Then DDD = DDD + (30 - old_day) + new_day

If old_mon = 7 Then DDD = DDD + (31 - old_day) + new_day

If old_mon = 8 Then DDD = DDD + (31 - old_day) + new_day

If old_mon = 9 Then DDD = DDD + (30 - old_day) + new_day

If old_mon = 10 Then DDD = DDD + (31 - old_day) + new_day

If old_mon = 11 Then DDD = DDD + (30 - old_day) + new_day

If old_mon = 12 Then DDD = DDD + (31 - old_day) + new_day

DDD = DDD + 1

Endif

`if more than 1 month separates the old and new dates...

If abs(new_mon - old_mon) > 1

for i = old_mon+1 to new_mon-1

if i = 2 then DDD = DDD + 28

if i = 3 then DDD = DDD + 31

if i = 4 then DDD = DDD + 30

if i = 5 then DDD = DDD + 31

if i = 6 then DDD = DDD + 30

if i = 7 then DDD = DDD + 31

if i = 8 then DDD = DDD + 31

if i = 9 then DDD = DDD + 30

if i = 10 then DDD = DDD + 31

if i = 11 then DDD = DDD + 30

next i

DDD = DDD - 1

Endif

`** MONTHS TO DAYS: if years are different **

`NOTE: we must do this step EVEN THOUGH there may be years in between the new and old years BECAUSE

`we need to find the DAYS REmAINING in that year AND the DAYS ELAPSED in the new year

IF abs(new_yr - old_yr) >= 1

`output

print "AT LEAST 1 yr difference"

`FIRST: use time differences to find the hours/seconds/minutes between the two times

SEC = (60 - old_sec) : old_min = old_min + 1

MIN = (60 - old_min) : old_hr = old_hr + 1

HRS = (24 - old_hr) : old_day = old_day + 1

`THEN: add the remaining seconds from the NEW TIMES

SEC = SEC + new_sec

MIN = MIN + new_min

HRS = HRS + new_hr

`THEN: find if the times have gone over the limits e.g. sec > 60 then must add one minute and reset by sec-60

If SEC >= 60 then SEC = SEC - 60 : MIN = MIN + 1

If MIN >= 60 then MIN = MIN - 60 : HRS = HRS + 1

If HRS >= 24 then HRS = HRS - 24 : DDD = DDD + 1

`Find the days remaining in that month

If old_mon = 1 Then DDD = DDD + (31 - old_day)

If old_mon = 2 Then DDD = DDD + (28 - old_day)

If old_mon = 3 Then DDD = DDD + (31 - old_day)

If old_mon = 4 Then DDD = DDD + (30 - old_day)

If old_mon = 5 Then DDD = DDD + (31 - old_day)

If old_mon = 6 Then DDD = DDD + (30 - old_day)

If old_mon = 7 Then DDD = DDD + (31 - old_day)

If old_mon = 8 Then DDD = DDD + (31 - old_day)

If old_mon = 9 Then DDD = DDD + (30 - old_day)

If old_mon = 10 Then DDD = DDD + (31 - old_day)

If old_mon = 11 Then DDD = DDD + (30 - old_day)

If old_mon = 12 Then DDD = DDD + (31 - old_day)

`Also find the days in the months between old month+1 and end of the year (12)

for i = (old_mon+1) to 12

if i = 2 then DDD = DDD + 28

if i = 3 then DDD = DDD + 31

if i = 4 then DDD = DDD + 30

if i = 5 then DDD = DDD + 31

if i = 6 then DDD = DDD + 30

if i = 7 then DDD = DDD + 31

if i = 8 then DDD = DDD + 31

if i = 9 then DDD = DDD + 30

if i = 10 then DDD = DDD + 31

if i = 11 then DDD = DDD + 30

if i = 12 then DDD = DDD + 31

next i

`find days in the months between start of NEXT year and the new_month - 1

for i = 1 to (new_mon-1)

if i = 1 then DDD = DDD + 31

if i = 2 then DDD = DDD + 28

if i = 3 then DDD = DDD + 31

if i = 4 then DDD = DDD + 30

if i = 5 then DDD = DDD + 31

if i = 6 then DDD = DDD + 30

if i = 7 then DDD = DDD + 31

if i = 8 then DDD = DDD + 31

if i = 9 then DDD = DDD + 30

if i = 10 then DDD = DDD + 31

if i = 11 then DDD = DDD + 30

next i

`Finally, add the days already elapsed in the new month

DDD = DDD + new_day

ENDIF

`if there's more than 1 year between the 2 dates

IF abs(new_yr - old_yr) > 1

print "MORE THAN 1 YEAR" : print ""

DDD = DDD + (((new_yr - old_yr)-1) * 365)

`** LEAP YEARS **

`basically, run a loop through ALL the years in between old and new years. run another loop within between 2012 and 5016 BUT step 4 years in between

`if the current year (i) = a leap year (given by "a") then increase the days by 1

For i = old_yr to new_yr

for a = 1904 to 3016 step 4

if i = a then DDD = DDD + 1  : print "LEAP YEAR" : print ""

next a

Next i

ENDIF

`** FINAL CONVERSION FOR OUTPUT **

FINAL = SEC + (MIN * 60) + (HRS * 60 * 60) + (DDD * 24 * 60 * 60)

`** Output **

Print "--------"

Print "FINAL OUTPUT"

Print "--------"

Print DDD ; " days"

Print HRS ; " hours"

Print MIN ; " minutes"

Print SEC ; " seconds"

Print ""

Print "FINAL:  " ; FINAL

endfunction FINAL

if memblock exist(1)=1 then delete memblock 1

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CodeBase - Duration calculator - time elapsed calculator

Information

Summary

Full Description

DarkBASIC Professional Source Code

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Author Name:	HWT
CodeBase Entries:	1