awk need epoch days logic for data transformation - awk

Question

In excel when I give in the cell A1=1/1/1970 and A2=today(), A2 resolves to 17-Apr-2020.
And when I subtract A2 with A1, I get 18369 which is the epoch days for my data transformation logic.
I'm trying to simulate the same functionality in awk, but not getting the whole number
$ awk ' BEGIN { t=strftime("%F",systime()); print t; gsub("-"," ",t); t2=mktime(t " 0 0 0" ); print t2/(24*60*60) } '
2020-04-17
18368.8
$
Is this translation for awk correct?. What is the reason for 0.2 difference in my awk code. How to fix it.

Answer 1

Converting my comment to answer so that solution is easy to find for future visitors.
You are seeing this difference due to timezone offset between your current local timezone and UTC.
On more recent gnu awk version you can use this to pass an additional utc-flag to mktime function:
awk 'BEGIN {t=strftime("%F",systime()); print t;
gsub("-"," ",t); t2=mktime(t " 0 0 0", 1); print t2/(24*60*60)}'
However if you are on older awk versions then use this work-around to get time computations in UTC:
TZ=UTC awk 'BEGIN {t=strftime("%F",systime()); print t;
gsub("-"," ",t); t2=mktime(t " 0 0 0"); print t2/(24*60*60) } '

Answer 2

The epoch days involve calculating how many multiples of "86400" seconds have passed since 1-Jan-1970. We have a boundary issue in this calculation, i.e we should not take 1st Jan, 1970 as 1 epoch day because the particular day has not completed.
That is the reason the java epoch days function returns 0 for 1-Jan-1970
java.time.LocalDate.parse("1970-01-01").toEpochDay
res2: Long = 0
Now, with this for 17-Apr-2020, we get
java.time.LocalDate.parse("2020-04-17").toEpochDay
res3: Long = 18369
Let's try to get the same answer using awk.
awk -v dt="2020-04-17" '
function epoch_days(dta)
{
yr=dta[0];
for(i=1970;i<strftime("%Y",mktime(dta[1] " 12 31 0 0 0"));i++)
e1+=strftime("%j",mktime(i " 12 31 0 0 0"))+0;
e1+=strftime("%j",mktime(dta[1] " " dta[2] " " dta[3] " 0 0 0"))-1;
return e1
}
BEGIN {
split(dt,dta,"-");
print epoch_days(dta) ;
}'
18369
Lets cross verify for current date i.e 22-Aug-2021
java.time.LocalDate.now
res5: java.time.LocalDate = 2021-08-22
java.time.LocalDate.now.toEpochDay
res6: Long = 18861
And our awk code returns
awk '
function epoch_days(dta)
{
yr=dta[0];
for(i=1970;i<strftime("%Y",mktime(dta[1] " 12 31 0 0 0"));i++)
e1+=strftime("%j",mktime(i " 12 31 0 0 0"))+0;
e1+=strftime("%j",mktime(dta[1] " " dta[2] " " dta[3] " 0 0 0"))-1;
return e1
}
BEGIN {
dt=systime(); # 22-Aug-2021
dt2=strftime("%F",dt);
split(dt2,dta,"-");
print epoch_days(dta) ;
}'
18861