How to parse only selected column values using awk

Since no sample output is shown, I don't know if this output is what you want. It is the output one gets from the code provided with the minimal changes required to get it to run:

$ awk -v k='3 5 8' 'BEGIN{split(k,a," ");} {for(i=1;i<=length(a);i++){print $a[i]}}' testUnix.txt 
array
array
array



SA
AUS
ARZ

The above code prints out the selected columns in the same order supplied by the variable k.

Notes

• The awk code never defined ksh_arr2. I presume that the value of this array was to be passed in from the shell. It is done here using the -v option to set the variable k to the value of ksh_arr2.

• It is not possible to pass into awk an array directly. It is possible to pass in a string, as above, and then convert it to an array using the split function. Above the string k is converted to the awk array a.

• awk syntax is different from shell syntax. For instance, awk does not use do or done.

Details

• -v k='3 5 8'

  This defines an awk variable k. To do this programmatically, replace 3 5 8 with a string or array from the shell.

• BEGIN{split(k,a," ");}

  This converts the space-separated values in variable k into an array named a.

• for(i=1;i<=length(a);i++){print $a[i]}

  This prints out each column in array a in order.

Alternate Output

If you want to keep the output from each line on a single line:

$ awk -v k='3 5 8' 'BEGIN{split(k,a," ");} {for(i=1;i<length(a);i++) printf "%s ",$a[i]; print $a[length(a)]}' testUnix.txt 
array array array

SA AUS ARZ

awk 'NR>=1 { print $3 " " $5 " " $8 }' testUnix.txt