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Substitute string with an index number

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# 1  
Old 06-19-2013
Substitute string with an index number
Objective is to substitute Jan with 01, Feb with 02 and so on. The month will be provided as input.

I could construct below awk and it worked.
Code:
echo Jun | \
awk 'BEGIN{split("Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec",mon," ")}{ for (i=1;i<=12;i++){ if ($1==mon[i]) printf("%02d\n",i)} }'

Wanted to know if there are better alternatives
Last edited by vbe; 06-19-2013 at 09:11 AM..
# 2  
Old 06-19-2013
Code:
$ date -j -f "%b" "Jun" "+%m"
06

# 3  
Old 06-19-2013
@rajamadhavan - Did not work. I'm in AIX.
# 4  
Old 06-19-2013
AIX doest seems to have the format conversion options..

may need to do something like this..

Code:
echo "Jan" | awk 'BEGIN{a["Jan"]="01";a["Feb"]="02"}{print a[$0]}'

# 5  
Old 06-19-2013
Code:
$ echo Jun | awk ' { printf("%02d",int(index("JanFebMarAprMayJunJulAugSepOctNovDec",$1)/3)+1) } '
06

These 2 Users Gave Thanks to anbu23 For This Post:
elixir_sinari krishmaths
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