mjhjmtu

I want to print column 3 and 4 but without printing the headers empty column lines.

header
tap
norm 
X Y 880 1787
X Y 3253 3439

Printing this via awk 'print($3,$4)' gives

(BLANK lines from header rows)
880 1787
3253 3439

This works to print column 19 only but how would I get it to print both columns?

awk '$3print $3' 
880
3253

I want col3 col4

880 1787
3253 3439

I tried but get only column 3 printed

awk '$3print $3''$4print $4' 
880
1787
3253
3439

You know how long the head is (three lines):

awk 'NR > 3 print $3, $4 '

NR is the number of records (lines by default) that has been read so far. With NR > 3 we trigger the print statement for the fourth line and all lines after.

You could use

awk '$3 print $3, $4 '

but this is a bit fragile as it depends on not only $3 being present but also that it is non-zero (a zero in the third column would give no output for that line).

If

header
tap
norm
X Y 880 1787
X Y 3253 3439

Then

awk 'print $3 " " $4' | sed '/^ *$/d'

Makes

880 1787
3253 3439

And for your last part you can try

 awk 'print $3 " " $4 " " $19'

To print the 3rd, 4th and 19th columns.