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Capturing Data between first quote and next quote


 
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# 1  
Old 09-26-2008
Capturing Data between first quote and next quote

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  
Old 09-26-2008
Code:
awk ' if (index($0, "'") >0 ) {split($0,arr, "'") ;print arr[2]} else {continue}' file

# 3  
Old 09-26-2008
Hi.

An sed directive with regular expressions can be used, but may be more cryptic:
Code:
#!/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:
Code:
% ./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  
Old 09-26-2008
Code:
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  
Old 09-26-2008
if you have Python:
Code:
#!/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


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