if grep -- don't want it to print the matched string

[GaijinPunch]

Greets -- I'm sure this is something incredibly silly and easy, that I'm just not getting. Let's say I've got this:

if grep "nimrod" mylog.log
then
print "there's a nimrod in your log!"
fi

And w/ a log file like
Bob is kewl
Jane is an idiot
You are a nimrod

the out put would be
$>nimrod.ksh
You are a nimrod
there's a nimrod in your log!

I want to get rid of the string that's matched in the log file, and only print out the string in the print statement when a match is found. Easy?

Thanks

 

[GaijinPunch]

Well, I guess I could output to a file, and do another grep w/ the -v option. Anything that might save that step?

 

[tdatgod]

Hi,
the status is set regardless of whether the output is displayed. Just put the output from the GREP to /dev/null which is the bit bucket. then check that status $?.

however I can't remember if GREP returns 0 when it works so you might have to negate the status.


grep "nimrod" mylog.log > /dev/null
if [ ! $? ]
then
print "there's a nimrod in your log!"
fi

 

[GaijinPunch]

Good deal man. I'll give that a shot.
Thanks!

 

[aigles]

You can use the -q option of grep.
With this option grep do not write anything to stdout and exit with zero status if an input line is selected.

grep -q "nimrod" mylog.log && print "there's a nimrod in your log!"

Jean Pierre.

 

[GaijinPunch]

Thanks aigles -- that's what I was looking for!

 

[rarnold1]

For this type of operation would write as follows

GREPCOUNT=`grep -c "nimrod" mylog.log`
if [[ $GREPCOUNT > 0 ]]
then
print "there's a nimrod in your log!"
fi