how to extract path from string in shell script

Question

i need to extract path from string

for e.g

title="set key invert ; set bmargin 0 ; set multiplot ; set size 1.0 , 0.33 ; set origin 0.0 , 0.67 ; set format x "" ; set xtics offset 15.75 "1970-01-01 00:00:00" , 31556736 ; plot "/usr/local/lucid/www/tmp/20171003101438149255.dat" using 1:5 notitle with linespoints ls 2'"

Then expected output should be

/usr/local/lucid/www/tmp/20171003101438149255.dat

using awk or grep

Are you asking for help to parse a file containing that text, or a variable named title set as you show (which then would be wrong since you're trying to contain " within a "-delimited string) or something else? — Ed Morton
– Ed Morton, Commented Oct 3, 2017 at 13:36

RomanPerekhrest · Accepted Answer · 2017-10-03 09:44:57Z

0

sed approach:

title='set key invert ; set bmargin 0 ; set multiplot ; set size 1.0 , 0.33 ; set origin 0.0 , 0.67 ; set format x "" ; set xtics offset 15.75 "1970-01-01 00:00:00" , 31556736 ; plot "/usr/local/lucid/www/tmp/20171003101438149255.dat" using 1:5 notitle with linespoints ls 2'

sed 's/.* plot "\([^"]\+\).*/\1/' <<<$title
/usr/local/lucid/www/tmp/20171003101438149255.dat

answered Oct 3, 2017 at 9:44

RomanPerekhrest

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CWLiu · Accepted Answer · 2017-10-03 09:51:07Z

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With grep solution,

grep -oP '"\K/[^"]*(?=")' <<< $title

With awk solution,

awk '{match($0,/\/[^"]*/,a);print a[0]}' <<< $title

edited Oct 3, 2017 at 9:51

answered Oct 3, 2017 at 9:45

CWLiu

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P.... · Accepted Answer · 2017-10-03 10:19:00Z

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Shorter regex with grep:

grep -oP 'plot "\K[^"]+' <<< $title
/usr/local/lucid/www/tmp/20171003101438149255.dat

answered Oct 3, 2017 at 10:19

P....

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