convert month from Aaa to xx in little script with awk

Question

I am trying to report on the number of files created on each date. I can do that with this little one liner:

ls -la foo*.bar|awk \'{print $7, $6}\'|sort|uni         


        

Answer 1

Here's the idiomatic way to convert an abbreviated month name to a number in awk:

$ echo "Feb" | awk '{printf "%02d\n",(index("JanFebMarAprMayJunJulAugSepOctNovDec",$0)+2)/3}'
02

$ echo "May" | awk '{printf "%02d\n",(index("JanFebMarAprMayJunJulAugSepOctNovDec",$0)+2)/3}'
05

Let us know if you need more info to solve your problem.

Answer 2

Just use the field number of your month as an index into the months array.

print months[$6]

Since ls output differs from system to system and sometimes on the same system depending on file age and you didn't give any examples, I have no way of knowing how to guide you further.

Oh, and don't parse ls.

Answer 3

Assuming the name of the months only appear in the month column, then you could do this:

ls -la foo*.bar|awk '{sub(/Jan/,"01");sub(/Feb/,"02");print $7, $6}'|sort|uniq -c

Answer 4

To parse AIX istat, I use:

istat .profile | grep "^Last modified" | read dummy dummy dummy  mon day time dummy yr dummy
echo "M: $mon D: $day T: $time Y: $yr"
-> Month: Mar Day: 12 Time: 12:05:36 Year: 2012

To parse AIX istat month, I use this two-liner AIX 6.1 ksh 88:

monstr="???JanFebMarAprMayJunJulAugSepOctNovDec???"
mon="Oct" ; hugo=${monstr%${mon}*} ; hugolen=${#hugo} ; let hugol=hugolen/3 ; echo "Month: $hugol"
-> Month: 10

1..12 : month name ok

If lt 1 or gt 12 : month name not ok

Instead of "hugo" use speaking names ;-))