bash script and grep

syros · 01-13-2005, 09:48 AM

Hi everyone, i have to do a little script that search files that contains a string, and print the name of thoose files on the screen. Also, it must allow the user to see only the x first or last results. The script is already done and works well, but it is giving a strange message when asking only for the first x results.

For this cas i use this command:

Code:

find /usr/include/ -name "*" -print | xargs grep -s -l $1 |  head -n $3

so when i run

Code:

 ./script.sh class + 3

to get the 3 first results i get this:

Code:

/usr/include/awk/regex.h
/usr/include/awk/regex_internal.h
/usr/include/net/route.h
xargs: grep: terminated by signal 13

Does anyone know what i have to do so that the xargs message doensn't appear?

It seems it happens cause head ends the grep process.

Well, thanks in advance.

homey · 01-13-2005, 10:07 AM

Quote:

find /usr/include/ -name "*" -print | xargs grep -s -l $1 | head -n $3

Try changing $3 to just 3

bigearsbilly · 01-13-2005, 10:14 AM

incidentally you don't need -name '*'

try:

Code:

blah/blah blah |  2>/dev/null xargs grep | head

though i don't like this solution much.

interesting point though.

syros · 01-13-2005, 10:21 AM

Many thanks, that worked perfectly!

jlliagre · 01-13-2005, 03:04 PM

incidentally, you do not need "-print" either ...