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regexbashshquotes

Replace all commas between two quotes in a bash script

I need that all "," between two " are replaced with ";" within a bash script. I'm close, but hours on the internet and stackoverflow led me to this:

echo ',,Lung,,"Lobular, each.|lungs, right.",false,,,,"organ, left.",,,,,' | sed -r ':a;s/(".*?),(.*?")/\1;\2/;ta'

With the result:

,,Lung,,"Lobular; each.|lungs; right.";false;;;;"organ; left.",,,,,

Correct would be:

,,Lung,,"Lobular; each.|lungs; right.",false,,,,"organ; left.",,,,,
Solution

  • Not sure how you want to deal with lines that have an odd number of double quotes (eg, the double quoted string spans multiple lines), but perhaps:

    awk '!(NR%2){gsub(",",";")} 1' RS=\" ORS=\"

    This simply treats " as the record separator and does the replacement only on odd numbered records. Seems to work as desired. (Or, rather, it works as you seem to desire!)

    As oguz points out in a comment, this prints an extra " at the end. That can be fixed with:

    awk '!(NR%2){gsub(",",";")} {printf RFS $0} {RFS="\""}' RS=\"

    which is a bit uglier but more correct . (or, rather, less incorrect!) If your input stream ends with a ", that quote will be truncated. If, however, your input is terminated by a newline rather than a ", this will do what you want.

    OTOH, you might just want to do:

    perl -wpE 'BEGIN{$/=\1}; y/,/;/ if $in; $in = ! $in if $_ eq "\""'

    Which reads one character and uses a simple state machine. ($_ is the current character, so $in = ! $in changes state when a double quote is seen and the transliteration only happens when $in is non-zero.)