finding the line number from a grep ?

Shell Programming and Scripting
1

Hi there

does anybody know how i can get the line number from an entry or entries in a file ?? for example if i had a file

test1
test2
test3
test1

and i needed to get the line numbers for all instances of test1 in that file

with the answer being (1,4)

Would anybody be able to point me in the right direction with this ..?

Any help would be greatly appreciated

2 
$ cat samplefile
test1
test2
test3
test1

$ grep -n test1 samplefile
1:test1
4:test1
3

An approach with awk:

awk '/test1/{s=s?s","NR:NR}END{print s}' file

Regards

4

Why not simply:

awk '/test1/ {print NR}' file

Or did I miss something in the OP?

Edit: Oh, I see. It's the desired output format (n,n)! Disregard my remark.

5

i have a file amit.txt

#cat amit.txt
amit
ars
amit
arg

i want to the patten amit to be serched.

The code is

=============
grep -n amit amit.txt >ars

sed 's/\(.\).*/\1/' ars

6 
awk '{
        if (_[$0]=="")
        _[$0]=NR
        else
        _[$0]=sprintf("%s,%s",_[$0],NR)
        }
        END{
        for (i in _)
                print i" ("_,")"
        }' a
7

This will meet your requirements but has too many pipes:D

$ grep -n test1 samplefile | awk -F":" '{print $1}' | tr "\n" ",\n" | sed 's/,$/\n/'
1,4
8

thanks guys