Capturing Data between first quote and next quote

Shell Programming and Scripting
1

I have input file like

RDBMS FALIURE UTY8703 'USER_WORK.TEST' .HIghest return code '12'

I want to parse data which comed between first quote till next quote

USER_WORK.TEST

can you please suggest how to do that

2 
awk ' if (index($0, "'") >0 ) {split($0,arr, "'") ;print arr[2]} else {continue}' file
3

Hi.

An sed directive with regular expressions can be used, but may be more cryptic:

#!/bin/bash -

# @(#) s2       Demonstrate extraction of string between left-most markers.

echo
echo "(Versions displayed with local utility \"version\")"
version >/dev/null 2>&1 && version "=o" $(_eat $0 $1) sed
set -o nounset
echo

FILE=${1-data1}

echo " Data file $FILE:"
cat $FILE

echo
echo " Results:"
sed -n "s/[^']*'\([^']*\)'.*/\1/p" $FILE

exit 0

Producing:

% ./s2

(Versions displayed with local utility "version")
Linux 2.6.11-x1
GNU bash 2.05b.0
GNU sed version 4.1.2

 Data file data1:
RDBMS FALIURE UTY8703 'USER_WORK.TEST' .HIghest return code '12'

 Results:
USER_WORK.TEST

See man sed for details ... cheers, drl

4 
using awk:
 
echo "RDBMS FALIURE UTY8703 'USER_WORK.TEST' .HIghest return code '12'" | awk -F"'" '{print $2}'
 
using cut:
 
echo "RDBMS FALIURE UTY8703 'USER_WORK.TEST' .HIghest return code '12'" | cut -d"'" -f2
5

if you have Python:

#!/usr/bin/env python
for line in open("file"):
    start=line.index("'") #check for position of first quote
    end=line[start+1:].index("'") # check for position of second quote
    print line[start+1:start+end] #get value