UNIX for Advanced & Expert Users

QUESTION: How can I cut out the date from just the first line and reformat it to 31-Jul-2007? I'll restate the question at the bottom again...

DESCRIPTION: I need to cut a date out of a file - an example of the date's format in the file is 2007-07-31. It's in the 5th field and is separated by y with an accent over it, so I need to print the value of that delimiter. There are 10's of thousands of lines in the file, but I only need to scan the first line for this date. So far I have the following, and while it at least gives me some kind of output, it's obviously not what I need (actually, it's returning every line in the file - I know this because it tried to do that before I copied it to a 'test' file and deleted all the other lines but the first):

DATE=`cut -d 'printf "\375"' -f5 < file.test`

echo $DATE returns the whole line (or really, the whole file), something like this (the y's have a forward accent over them):

18y835y000000ySy2007-06-13y09:40:38y

QUESTION: how can I cut out the date from just the first line and reformat it to 31-Jul-2007?

I found part of the answer:

cat tlog.test | cut -f5 -d`printf "\375"` | tail -1

Produces (I mentioned the wrong date above):

2007-06-13

Now how do I reformat it to 13-Jun-2007 without using a hard-coded date?

Code:

mVar='2007-06-11'
mYear=`echo $mVar | cut -d'-' -f1`
mMth=`echo $mVar | cut -d'-' -f2`
mDay=`echo $mVar | cut -d'-' -f3`
mAllMths='Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec'
mMthOut=`echo ${mAllMths} | cut -d' ' -f${mMth}`
echo ${mDay}'-'${mMthOut}'-'${mYear}