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parsing - How to find and print specific character in bash

I have file like:

AA,A=14,B=356,C=845,D=4516
BB,A=65,C=255,D=841,E=5133,F=1428
CC,A=88,B=54,C=549,F=225

I never know if in the row missing A,B,C or D value. But I need to transform this file like:

AA,A=14,B=356,C=845,D=4516,-,-
BB,A=65,-,C=255,D=841,E=5133,F=1428
CC,A=88,B=54,C=549,-,-,F=225

So if any value missing print just - mark. My plan is have the same number of columns to easy parsing. I am prefer awk solution. Thank you for any advice or help.

My first try was:

awk '{gsub(/[,]/, "")}; BEGIN{ FS = OFS = "" } { for(i=1; i<=NF; i++) if($i ~ /^ *$/) $i = "-" }; {print $0}'

But then I notice, that some values are missing.

EDIT:

From my header I know that there is value A,B,C,D,E,F...

See Question&Answers more detail:os

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$ cat file.txt
AA,A=14,B=356,C=845,D=4516
BB,A=65,C=255,D=841,E=5133,F=1428
CC,A=88,B=54,C=549,F=225

$ perl -F, -le '@k=(A..F);
   $op[0]=$F[0]; @op[1..6]=("-")x6;
   $j=0; for($i=1;$i<=$#F;){ if($F[$i] =~ m/$k[$j++]=/){$op[$j]=$F[$i]; $i++} }
   print join(",",@op)
   ' file.txt
AA,A=14,B=356,C=845,D=4516,-,-
BB,A=65,-,C=255,D=841,E=5133,F=1428
CC,A=88,B=54,C=549,-,-,F=225
  • -F, split input line on , and save to @F array
  • -l removes newline from input line, adds newline to output
  • @k=(A..F); initialize @k array with A, B, etc upto F
  • $op[0]=$F[0]; @op[1..6]=("-")x6; initalize @op array with first element of @F and remaining six elements as -
  • for-loop iterates over @F array, if element matches with @k array element in corresponding index followed by =, change @op element
  • print join(",",@op) print the @op array with , as separator

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