How to Grep of by section?

Shell Programming and Scripting
1

I have a script that outputs this as a file

John Smith
----------------
memberOf: example1;sampletest;test
memberOf: example2;sampletest;test
memberOf: example3;sampletest;test
memberOf: example4;sampletest;test
A Member of 4 Groups
 
Sally Smith
----------------
memberOf: example1;sampletest;test
memberOf: example2;sampletest;test
memberOf: example3;sampletest;test
A Member of 3 Groups
 
Toby Smith
----------------
memberOf: example1;sampletest;test
memberOf: example2;sampletest;test
memberOf: example3;sampletest;test
memberOf: example4;sampletest;test
memberOf: example5;sampletest;test
A Member of 5 Groups

I want to make it so that if some one uses (ex. grep Toby Smith) on the file, than they will get the name aswell as their memberships. Also, I can only use grep, so is there anything i can do in the script to allow grep to gather all the information for the each member.

2

Smash the group of lines into single lines and post process the grep to put it back.

Consider (assumes a Unix/Linux line terminated format in my case in a file called groups.txt):

sed -e 's/^$/^B/' <groups.txt |  tr '\012\002' '\011\012' | sed 's/^\o011//' | grep '^Sally Smith' | tr '\011' '\012'

You'll want the data to end with an empty line.

(oh and ^B is a Ctrl-B, I should have been more consistent in my approach)

3

Check this ,it may help you

awk  'BEGIN{RS="----------------"}/John/{print;getline;print}'  file
4

Hi.

An option in perl, -00, allows processing by paragraphs, provided they are separated by empty lines:

#!/usr/bin/env bash

# @(#) s1	Demonstrate processing paragraphs of lines, perl.

# Utility functions: print-as-echo, print-line-with-visual-space, debug.
# export PATH="/usr/local/bin:/usr/bin:/bin"
LC_ALL=C ; LANG=C ; export LC_ALL LANG
pe() { for _i;do printf "%s" "$_i";done; printf "\n"; }
pl() { pe;pe "-----" ;pe "$*"; }
db() { ( printf " db, ";for _i;do printf "%s" "$_i";done;printf "\n" ) >&2 ; }
db() { : ; }
C=$HOME/bin/context && [ -f $C ] && $C perl

PATTERN=${1-Sally}
FILE=data1

pl " Input data file $FILE, looking for \"$PATTERN\":"
cat $FILE

pl " Results:"
perl -00 -wn -e " print if m{$PATTERN}" $FILE

exit 0

producing:

$ ./s1

Environment: LC_ALL = C, LANG = C
(Versions displayed with local utility "version")
OS, ker|rel, machine: Linux, 2.6.26-2-amd64, x86_64
Distribution        : Debian 5.0.8 (lenny, workstation) 
bash GNU bash 3.2.39
perl 5.10.0

-----
 Input data file data1, looking for "Sally":
John Smith
----------------
memberOf: example1;sampletest;test
memberOf: example2;sampletest;test
memberOf: example3;sampletest;test
memberOf: example4;sampletest;test
A Member of 4 Groups

Sally Smith
----------------
memberOf: example1;sampletest;test
memberOf: example2;sampletest;test
memberOf: example3;sampletest;test
A Member of 3 Groups

Toby Smith
----------------
memberOf: example1;sampletest;test
memberOf: example2;sampletest;test
memberOf: example3;sampletest;test
memberOf: example4;sampletest;test
memberOf: example5;sampletest;test
A Member of 5 Groups

-----
 Results:
Sally Smith
----------------
memberOf: example1;sampletest;test
memberOf: example2;sampletest;test
memberOf: example3;sampletest;test
A Member of 3 Groups

This also allows constructs such as:

$ ./s1 'John|Toby'

producing:

 Results:
John Smith
----------------
memberOf: example1;sampletest;test
memberOf: example2;sampletest;test
memberOf: example3;sampletest;test
memberOf: example4;sampletest;test
A Member of 4 Groups

Toby Smith
----------------
memberOf: example1;sampletest;test
memberOf: example2;sampletest;test
memberOf: example3;sampletest;test
memberOf: example4;sampletest;test
memberOf: example5;sampletest;test
A Member of 5 Groups

Best wishes ... cheers, drl

5

How come that you can only use grep? Unfortunately, grep by itself does not have the functionality you are hoping for.

You make reference to a script. Could you post what you have been working on?