regexp to get first line of string - Shell Programming and Scripting

jonas.gabriel · #1 (**permalink**) 05-08-2007

Hi everybody

Code:

for file in *
#Bash performs filename expansion
#+ on expressions that globbing recognizes.
do
	
	output="`grep -n "$1" "$file"`"
	echo "$file: `expr "$output" : '\(^.*$\)'`"

done

In the above bash script segment, I try to print just the first line of string named output.

I've got a problem with the regexp. It returns the full string which is multilined.

Thank you in advance for any help

aigles · #2 (**permalink**) 05-08-2007

You can try :

Code:

echo ""$file: `echo "$output" | sed -n 1p`"

Jean-Pierre.

anbu23 · #3 (**permalink**) 05-08-2007

Code:

grep -n "$1" "$file" | read output

jonas.gabriel · #4 (**permalink**) 05-09-2007

Thank all of you

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