Print column contents by column name

Awk 1 liner for above problem (if you are interested):

awk -v col=COL2 'NR==1{for(i=1;i<=NF;i++){if($i==col){c=i;break}} print $c} NR>1{print $c}' file.txt

awk -v col=COL3 'NR==1{for(i=1;i<=NF;i++){if($i==col){c=i;break}} print $c} NR>1{print $c}' file.txt

Just pass your column name COL1, COL2, COL3 etc with -vcol= flag.

Note that the first solution prints out the whole file if the named column does not exist. To output a warning message if this occurs try

awk -v col=NoneSuch 'NR==1{for(i=1;i<=NF;i++){if($i==col){c=i;break}}   if (c > 0) {print $c}} else {print "Column " col "does not exist"} NR>1 && c > 0 {print $c}' file1.txt

a slight modification of anubhava post on top, for multiple columns

awk -vcol1="COL2" -vcol2="COL6" 'NR==1{for(i=1;i<=NF;i++){if($i==col1)c1=i; if ($i==col2)c2=i;}} NR>0{print $c1 " " $c2}' file.txt

when NR>1 does not print the column headers. This was modified to NR>0 which should print the columns with header names.