strip leading zeros in awk program

Question

So I\'m pretty lost as I missed a week and am playing catch up, but I\'m to write an awk program to tell the difference, in days, between two dates.

I\'m more or les

Answer 1

You need this, assuming your variable is called month:

gsub ("^0*", "", month);

The ^ is the start anchor and 0* means zero or more 0 characters. So this effectively removes all 0 characters at the start of the variable.

By way of example (which also shows a way to do it to the middle number as well, before splitting the date apart, see the second gsub for that):

pax> echo '1/1/2013
03/12/2014
02/02/1965' | awk '{gsub ("^0*", "", $0); gsub ("/0*", "/", $0); print}'

1/1/2013
3/12/2014
2/2/1965

Answer 2

Another option:

$ echo 01234 | awk '{x=$0+0;print x}'
1234

Answer 3

I'm sorry. My answer is not about awk, and it is about bash itself. So please don't give a minus for my comment)

echo -e "00000000" | sed -r 's/^[0]*$/0/g' | sed -r 's/^0[0]+//g'

Or perform integer calculations with (( )) notations

echo $((0000))

Answer 4

I tried the answer suggested by @paxdiablo. It ended up stripping the non-leading zeros as well.

$ echo '20/02/1965' | awk '{gsub ("^0*", "", $0); gsub ("/0*", "/", $0); print}'
2/2/1965

Using sub instead gsub worked fine for me.

$ echo '20/02/1965' | awk '{sub ("^0*", "", $0); sub ("/0*", "/", $0); print}' 
20/2/1965

Answer 5

Latest contribution somewhat shortened.

echo '20/02/1965' | awk '{sub ("/0", "/", $0)}1'
20/2/1965