Number of days between two dates

[kofa kofa]

Hello everyone
Happy New Year
I'm looking for a code that calculates the difference between an old date and today's date. For example, I want to calculate the number of days from April 20, 1922, to today's date.

[Bert Mariani]

Hello Kofa

See Ring Documentation:   Section 19.7

19.7 DiffDays() Function

Syntax:

DiffDays(cDate1,cDate2) ---> number of days (Date1 - Date2)

Example:

cDate1 = date()

see cDate1 + nl # 24/05/2015

cDate2 = adddays(cDate1,10)

see cDate2 + nl # 03/06/2015

see "DiffDays = " + diffdays(cDate1,cDate2) + nl # -10

see "DiffDays = " + diffdays(cDate2,cDate1) + nl # 10

[kofa kofa]

I tried many times, but it didn't work with old dates.

[kofa kofa]

Can you write the code that calculates the number of days from 20/4/1922?

[Mahmoud Fayed]

Hello Kofa

See this example:

Greetings,

Mahmoud

[Bert Mariani]

Hello Mahmoud,  Kofa

Looks like there is a Bug in DiffDays()

Has to do with the  Unix Epoch date January 1, 1970

OldDate..: 20/04/1922
Today....: 02/01/2026
DaysSince: 0

OldDate..: 20/04/1992
Today....: 02/01/2026
DaysSince: 12310

OldDate..: 20/04/1972
Today....: 02/01/2026
DaysSince: 19615

OldDate..: 20/04/1965
Today....: 02/01/2026
DaysSince: 0

[Bert Mariani]

DiffDays()  Bug

OldDate..: 01/01/1970
Today....: 02/01/2026
DaysSince: 20455

OldDate..: 31/12/1969
Today....: 02/01/2026
DaysSince: 0

[kofa kofa]

[Mahmoud Fayed]

Hello Bert, Kofo

Diffdays() is based on mktime() C function

On Windows, the range is 1970-2038

So, please QDate as in the next example

Source Code

load "guilib.ring"

d1 = new QDate() { setDate(2026,01,02) }

? d1.day()

? d1.month()

? d1.year()

d2 = new QDate() { setDate(1922,04,20) }

? d2.day()

? d2.month()

? d2.year()

? d1.daysto(d2)

Greetings,

Mahmoud

[kofa kofa]

thank you very much

[Mahmoud Fayed]

Hello Kofa

You are welcome :D

Greetings,

Mahmoud

[Bert Mariani]

An Alternative using  EpochTime()

========================

load "stdlibcore.ring"
// EpochTime(Date(), Time() )  // return EpochSec

CurSec  = EpochTime(Date(), "00:00:00" )  
OldSec  = EpochTime("20/04/1922", "00:00:00" )
DiffSec = CurSec - OldSec

See "CurSec: "+ CurSec +nl
See "OldSec: "+ OldSec +nl

NbrDays = DiffSec / 86400
See "NbrDays: "+ NbrDays +nl

=========================

CurSec: 1767312000        
OldSec: -1514764800       
NbrDays: 37987

========================

[kofa kofa]

the result not change if you change  for example : 10/10/1922

[kofa kofa]

[Mahmoud Fayed]

Hello Kofa, Bert

EpochTime() implementation uses DiffDays()

So, it's better to use QDate to avoid the (1970 to 2038 limitation on Windows)

Greetings,

Mahmoud

[Mahmoud Fayed]

Hello Bert

>> "OldSec  = EpochTime("20/04/1922", "00:00:00" )"

Started from Ring 1.25 this will produce an error message on Windows (by DiffDays() called from EpochTime())

As a result of this update (done today), where we detect this error: Update Visual Source (Using PWCT) - Revise diffdays() implementation · ring-lang/ring@81af280

By checking if vTimer == -1

Greetings,

Mahmoud

[kofa kofa]

thanks

[Mahmoud Fayed]

Hello Kofa

You are welcome :D

Greetings,

Mahmoud

[Bert Mariani]

Hello Mahmoud , Kofa

Another method

Using Astronomical days based on Julian Period

The Julian period is a chronological interval of 7980 years

### Astronomers refer to a Julian date as the number of days
### since the beginning of the Julian Period (January 1, 4713 BC at 12:00 Noon).
###
### 2026-01-02 = 2,461,043 days  
### 1922-04-20 = 2,423,165 days
### Difference =    37,878 days

========================

load "stdlibcore.ring"

CurYYYY = 2026  CurMM = 01  CurDD = 02
OldYYYY = 1922  OldMM = 04  OldDD = 20

? "CurDate: " + CurYYYY +","+ CurMM +","+ CurDD
? "OldDate: " + OldYYYY +","+ OldMM +","+ OldDD
? ""

CurJulian = JulianDay(CurYYYY,CurMM,CurDD)
OldJulian = JulianDay(OldYYYY,OldMM,OldDD)

NbrDays   = CurJulian - OldJulian

? "CurJulian: " + CurJulian
? "OldJulian: " + OldJulian
? "NbrDays..: " + NbrDays


# ---------- Time / Calendar ----------
func JulianDay(y,m,d)
    yy = y
    mm = m
    if mm <= 2
        yy = yy - 1
        mm = mm + 12
    end
    A = floor(yy/100)
    B = 2 - A + floor(A/4)
    jd = floor(365.25*(yy + 4716)) + floor(30.6001*(mm + 1)) + d + B - 1524.5

    return jd
end

# =================================

OUTPUT

CurDate: 2026,1,2
OldDate: 1922,4,20

CurJulian: 2461042.50
OldJulian: 2423164.50
NbrDays..: 37878

[Mahmoud Fayed]

Hello Bert

Thanks for sharing :D

Greetings,

Mahmoud

[kofa kofa]

thank you