How to print column based on row number

Shell Programming and Scripting
1

Hi,
I want to print column value based on row number say multiple of 8.

Input file:

line 1 67  34
line 2 45  57
. . .
. . . 
line 8 12  46
. . .
. . .
line 16 24  90
. . .
. . .
line 24 49  67

Output

46
90
67

I tried

"for ((i = 1; i<=3; i++)) 
do
b=$(echo "scale=2;($i*8)"| bc)
awk "NR==$b {print $2}" input file>> output file
done"

But it prints both the columns. I want to print only 2nd column.
Please help

2

Use single quotes instead of double quotes:

awk "NR==$b"' {print $2}' ....

Otherwise $2 gets expanded by the shell.

You could do this all in one awk like this:

awk '!(NR%8){print $2}' infile
3

I works. Thank you