Bash select multiple values in CSV column

Question

I have the following CSV format:

"/opt=920MB;4512;4917;0;4855","/=4244MB;5723;6041;0;6359","/tmp=408MB;998;1053;0;1109","/var=789MB;1673;1766;0;1859","/boot=53MB;656;692;0;729"

I would like to extract 2 values from each column, the first and the last value from the array, after the "=", like this:

"/opt=920MB;4855","/=4244MB;6359","/tmp=408MB;1109","/var=789MB;1859","/boot=53MB;729"

Per column it can work like this:

echo "$string" | awk 'BEGIN{FS=OFS=";"} {split($0,a,";"); print a[1],a[5]}'
/opt=920MB;4855

Any hints are highly appreciated.

What's up with your previous question - stackoverflow.com/q/51250515/5291015. Didn't that work out? — Inian
– Inian, Commented Jul 10, 2018 at 11:28
This is different from that one, this is to get 2 values from each column, not only one. — Constantin
– Constantin, Commented Jul 10, 2018 at 11:47

Teemv Teeccary · Accepted Answer · 2018-07-10 11:33:24Z

0

s/\([^;]*\);\([^;]*\)\[^"]*\([^;]*\);\([^;]*\)\[^"]*\([^;]*\);\([^;]*\)\[^"]*\([^;]*\);\([^;]*\)\[^"]*\([^;]*\);\([^;]*\)\[^"]*/\1;\2;\3;\4;\5;\6;\7;\8;\9;\10/

answered Jul 10, 2018 at 11:33

Teemv Teeccary

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kvantour Over a year ago

Hi and welcome to Stack Overflow. While this answer might be amazing, could you please elaborate on it and explain a bit why this answers the question.

Constantin · Accepted Answer · 2018-07-10 12:54:03Z

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Found the solution by using sed:

echo "$string"| sed 's/=*;[^"]*;/,/g'

answered Jul 10, 2018 at 12:54

Constantin

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Comments

Hasturkun · Accepted Answer · 2018-07-10 15:47:53Z

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You can do it without sed that way :

xargs -d"\"" -n 1 -a file.csv | xargs -d";" | awk '/ key (start|stop) / {next} {if (match($0,"/")) printf("\"%s;%s\",",$1,$NF)} END {print ""}' |  rev | cut -c 2- | rev

edited Jul 10, 2018 at 15:47

Hasturkun

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answered Jul 10, 2018 at 13:32

lue

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