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String between quotes

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# 1  
Old 05-08-2012
String between quotes
Hi,

Need to capture a string between 1st & last quote.

String can be anything like
a b "c" d e
Output: c
a "b c" d e
Output: b c
a "b c" d "e" f
Output: b c d e

sed 's/.*"\(.*\)".*/\1/g'
Above helps me to find the string between last quote.
Need to find the string between 1st & last quote.
# 2  
Old 05-08-2012
Code:
echo 'a "b c" d "e" f' | sed 's/[^"]*\"\(.*\)\".*/\1/; s/\"//g'

# 3  
Old 05-08-2012
Try like...

Code:
echo 'a b "c" d e ' | grep -Po '".*?"'

# 4  
Old 05-08-2012
Goodie
Lemme try to get this.

[^"]*
Takes care of all character apart from "

sed (.*) starts matching from backward
\"\(.*\)\".*

So the 1st segment tells there shouldn't be any " before that.

Am i right?
# 5  
Old 05-08-2012
If you are in ksh then you can slice the strings with the following:-
Code:
#!/bin/ksh

MYSTRING="a b c d e"            # Quoting to set the variable only. Quotes do not form part of string
TEMPVAR="${MYSTRING#*\"}" # Cut off everything before first quote - no effect
OUTPUT="${TEMPVAR%\"*}"   # Cut off everything after last quote - no effect
echo $OUTPUT

MYSTRING="a b \"c\" d e"            # Escaped quotes do form part of string
TEMPVAR="${MYSTRING#*\"}" # Cut off everything before first quote
OUTPUT="${TEMPVAR%\"*}"   # Cut off everything after last quote
echo $OUTPUT

The last criteria is a bit awkward though. The output you suggest ignores the fact that your string may have embedded quotes in it. Do you not want to know these exist?

If not, then append | tr -d "\"" to the echo $OUTPUT statement.



I hope that this helps.

Robin
Liverpool/Blackburn
UK
Last edited by rbatte1; 05-08-2012 at 08:00 AM.. Reason: I forgot the signing off bit
# 6  
Old 05-08-2012
Bug
Code:
awk -f '"' '{$1=$NF=""}1' file

Last edited by Franklin52; 05-08-2012 at 08:17 AM.. Reason: Please use code tags
# 7  
Old 05-08-2012
Try

Code:
sed -n 's/^[^"]*"\(.*\)"[^"]*$/\1/p' <filename>|tr -d '"'
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