I'm trying to figure this out.

awk '{print $1","$10","$11","$12","$13,$14,$15,$16,$17,$18,$19}' <<< "$PASTE_1" > test.csv

I need to print the $1 $10 $11 $12 separated by comma then continue with $13 until the of the line, without comma separation. Since there are many blank spaces from $13.

  • adding a sample input and output would help...
    – Sundeep
    Apr 24, 2017 at 9:47

3 Answers 3


Do you mean something like this:

awk '{a = ""; for (i = 13 ; i <= NF ; i++) a = a $i; 
       print $1 "," $10 "," $11 "," $12 "," a}'

The input

a b c d e f g h i j k l m n o p q r s t u v w x y z



That is, the fields starting from 13 are concatenated together, and then printed after 1, 10, 11 and 12.


One way to do it:

awk -v OFS=, '{print $1, $10, $11, $12, ($13 $14 $15 $16 $17 $18 $19) }'
perl -lane 'print join(",", @F[0,9..11], q//), @F[12..$#F]'


  1. Array @F holds the fields in a line. @F[...] refers to a slice of elements of array @F. So, @F[0,9..11] is the slice consisting of the 1st, 10th to 12 elements of @F which are then join-ed together by comma and a dummy empty element q// is suffixed to the slice to generate the required , for the coming slice. The other slice @F[12..$#F] gathers the 13th till the last element of the @F array via the $#F which holds the last element's idx. (Please note that array indexing is zero-starting in Perl).
  2. The various arguments to print are now joined together with the OFS = $, (defaults to null).

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