Shell Programming and Scripting

I have a value in a file i am processing that has a date like "2012-Jun-13"

how can I convert a date like that 2012-06-13?

Am I stuck building an array of three digit months and corresponding numbers and running through the logic of figuring out the number??

or can I convert this with strftime? if so I cant figure out how... Im also cool with dropping down to the shell level and using gnu date to do this. But I dont know how to pull a system call back into gawk as a variable.

Any ideas?

---------- Post updated at 10:26 AM ---------- Previous update was at 10:17 AM ----------

I found another post details what I need to do this with a system call..

If anyone knows how to do this without a system call, i would still like to know.

Thanks,
Trey

Code:

echo "2012-Jun-13" | awk -F- 'BEGIN {
	a["Jan"] = "01";
	a["Feb"] = "02";
	a["Mar"] = "03";
	a["Apr"] = "04";
	a["May"] = "05";
	a["Jun"] = "06";
	a["Jul"] = "07";
	a["Aug"] = "08";
	a["Sep"] = "09";
	a["Oct"] = "10";
	a["Nov"] = "11";
	a["Dec"] = "12";
} { print $1 FS a[$2] FS $3 }'