grep string and find line before

Shell Programming and Scripting
1

hi,

i have to grep for string in file but i want to find the group of this line so i must get lines before and select the group.
the file look like :

####name_groupe1
alphanumeric line
alphanumeric line
..
####name_groupe2
alphanumeric line
alphanumeric line
..
####name_groupe3
alphanumeric line
alphanumeric line
..

al the file is like that :

i want to grep string if i find it in line i want to gret the name_group

i use :

grep -n "string to search" file | cut -f1 -d: to get the line number but i want to go lines before to find the group and store that in variable.

i use script shell.

thanks

2

How about awk?

awk ' {
         /^###/ { line=NR; value=$0}
         if(index($0, "string to search")>0) { print line, value}
         } '  inputfilename
3

Or:

awk '$0 ~ str{print b}{b=$0}' str="String to find" file

Regards

4

Hi,

To be honest, i am not quiet sure about your req. I suppose it is as follow, please try it and find whether any help for you.

INPUT:

name_group1
a
a2
A
sdlfk
34
name_group2
b
b1
dfjkl
e4r
name_group3
dflk
line
this is a good idea

OUTPUT:

name_group1/name_group2/name_group3 according to your input string to be searched

CODE:

echo "Input string"
read str
nawk -v s="$str" 'BEGIN{
n=""
print s
}
{
if ($0 ~ /name/)
{
	n=$0
}
if (index($0,s)!=0)
{
	if (n!="")
		print n
	n==""
}
}' filename
5

hi,

summer_cherry you understand well my problem but i want to use awk not nawk because it's not supported in linux only in sun can u please try to help me to find the right solution using awk and storing the group_name in variable that i can use after.

thank u

:slight_smile: