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#1
07-31-2007
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Cut date using y hat as delimiter
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? Last edited by tekster757; 07-31-2007 at 09:24 AM. Reason: I found part of the answer 
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#2
07-31-2007
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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}
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