extract pattern data using gawk/awk

Question

I have a input text like below:

 10779: (255,255,  0,255) #FFFF00 yellow
   338: (255,238,125,255) #FFEE00 srgba(255,238,125,1)
   180: (238,221,  0,255) #EEDD00 srgba(238,221,0,1)
 ....

I wish the output is

 255,255,0,255    
 255,238,125,255    
 238,211,0,255

How to extract this data from the input using gawk/awk? I've tried this:

gawk 'BEGIN {FS=\"[:,()]\"};{print $3,$4,$5,$6}'

but this gives me space between the data, e.g. 255 255 0 255

You might find it easier to understand how to do this using regular expressions in a language like Perl, Python, Ruby, etc. Are you open to a solution in a language other than awk? — John Zwinck, Apr 05 '14 at 03:37
Set `OFS=","` so that fields are separated by commas on output. — Jonathan Leffler, Apr 05 '14 at 03:46

score 3 · Answer 1 · answered Apr 05 '14 at 03:45

3

Based on your input I am guessing this might work

awk -F'[()]' '{gsub(/ /,"",$2); print $2}' file
255,255,0,255
255,238,125,255
238,221,0,255

answered Apr 05 '14 at 03:45

iruvar

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it didn't went through – realmq Apr 05 '14 at 04:11
invalid char ''' in expression – realmq Apr 05 '14 at 04:15
@emilywuq - sure you running gawk? – iruvar Apr 05 '14 at 04:18
sorry, you are right, I guess it is awk on my windows not configured right. Can you explain the gsub part? – realmq Apr 05 '14 at 04:21
@emilywuq, in this case `gsub` does a global substitution of space chars with empty chars in field 2 – iruvar Apr 05 '14 at 04:24
You do not need the `$2` since `gsub`removes all the spaces, so this is ok: `awk -F'[()]' '{gsub(/ /,""); print $2}'` – Jotne Apr 05 '14 at 05:41
@jotne, you nitpicker ;-) All said and done you're an `awk` champ – iruvar Apr 05 '14 at 05:54
@1_CR I always try to get my `awk` as short and exact as possible ;) – Jotne Apr 05 '14 at 05:56
@jotne, short and exact as possible sounds suspiciously like `sed` ;-). In any case, thanks for keeping us journeymen `awk` practitioners honest – iruvar Apr 05 '14 at 05:59
@1_CR `sed` is god for lots of stuff, but it tends to be very hard to read and easy to miss a `.`. That is why I like `awk`, but correct tool for correct job :) – Jotne Apr 05 '14 at 06:12

Vijay · Answer 2 · 2014-04-05T04:26:50.287

1

Using perl:

perl -lne 'print $1 if(/\((.*?)\)/)'

If you donot want those spaces, then:

perl -lne '$a=$1 if(/\((.*?)\)/);$a=~s/ //g;print $a'

Awk:

awk '{a=index($0,"(");b=index($0,")");a++;b--;print substr($0,a,b-a)}'

Tested

edited Apr 05 '14 at 04:26

answered Apr 05 '14 at 04:11

Vijay

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how to do it using awk/gawk? – realmq Apr 05 '14 at 04:18
for removing the spaces you can anyhow use the gsub part by storing the substr in a string and then remove the spaces. – Vijay Apr 05 '14 at 04:28

jaypal singh · Answer 3 · 2014-04-05T04:22:20.217

1

Another way with awk:

$ awk -F'[(#]' 'gsub(/[ )]/,"")&&$0=$2' file
255,255,0,255
255,238,125,255
238,221,0,255

edited Apr 05 '14 at 04:22

answered Apr 05 '14 at 04:15

jaypal singh

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I want to get rid of space before 0 – realmq Apr 05 '14 at 04:16

mklement0 · Answer 4 · 2014-04-05T04:44:33.780

1

A sed solution, for variety:

sed -E 's/^.+\(([^)]+)\).+$/\1/; s/ //g' file

Or, one step at a time, with cut and tr:

cut -d '(' -f 2 file | cut -d ')' -f 1 | tr -d ' '

edited Apr 05 '14 at 04:44

answered Apr 05 '14 at 04:22

mklement0

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alas, it is still producing a different result than awk-based solutions (with the second line using GNU sed) – iruvar Apr 05 '14 at 04:37
@1_CR: Thanks for troubleshooting; fixed now. – mklement0 Apr 05 '14 at 04:44
1

+1, unlike the awk solutions, it leaves non-qualifying lines untampered – iruvar Apr 05 '14 at 04:46

extract pattern data using gawk/awk

4 Answers4