Search Results

Try:
awk -F'[ \t,_:.]*' '{print $5","$9":"$10":"$11"."$15}' infileawk -F'[ \t,_:.]*' '{print $5","$9":"$10":"$11"."$12","$15}' infile

sed 's/\(^\|[^[:alnum:]]\)\(rich\|RICH\|Rich\)\([^[:alnum:]]\|$\)/\1NICK\3/g' infile

# ./justdoit
1) one
2) two
3) three
4) four
5) five
6) quit
Please input your choice..1

This is f1 function

1) one
2) two
3) three
4) four
5) five
6) quit
Please input your...

That's a good point.

for i in $choice
do
echo "run function $i"
f${i}
done

should do the things.

Try something like:


f1 () {
echo running fn1
}

f2 () {
echo running fn2
}

f3 () {
echo running fn3
}

f4 () {
echo running fn4
}

echo "enter choice -- multiple...

sed 's/\(Person\(id=\|:\|ID:\)\).*/\1XXXXXX/' infile

For second question,you need provide some sample inputs and expect output first.

Does it have to be sed?
perl -pe 's/(?<![a-z])jo(?![a-z])/bo/ig' file

sed 's/user=.*/user=xxx/'
and:
sed 's/^am$/jo/i'

Um, no!

Sorry my mistake. I've updated my previous post :o

Hi.

Then you can either use ed, or write to a temporary file:


function Replace
{
ls | while read f; do
sed 's/JAM/BUTTER/g;s/BREAD/CRACKER/g;s/SCOOP/FORK/g;s/SPREAD/SPLAT/g' "$f" >...

Sorry, 1 is one day, so you need 1/24:

perl -MFile::Find -le'
find {
wanted => sub {
-f and 1 / 24 >= -M and print $File::Find::name;
}
}, shift
' ...

function search
{
cd $logsloc
echo "Enter date in format YYYYMMDD (enter to exit):"
read dateToLookFor
echo $dateToLookFor | grep -q ...