How to Remove Code Specific Code Lines using Unix - unix

Question

Could someone please help/advise how could I removed the first 4 line and the last 2 line of codes in my 3 JavaScript files using the Shell Script?
I tried using this guide: UNIX - delete specific lines but it will only work for the first 4 lines. All 3 Javascript files have different set of line of codes.
set -vx
lines2del="(1,2,3,4)"
sedCmds=${lines2del//,/d;}
sedCmds=${sedCmds/(/}
sedCmds=${sedCmds/)/}
sedCmds=${sedCmds}d
sed -i "$sedCmds" file
Any inputs are highly appreciated. Thanks

Answer 1

This might work for you (GNU sed):
sed -i '1,4d;N;$d;P;D' file
This deletes the lines 1 to 4 and then prints all other lines except the last two which it also deletes.

Answer 2

Add the following to your lines2del:
$(($(cat file | wc -l)-2)) // third last line
$(($(cat file | wc -l)-1)) // second last line
$(cat file | wc -l) // last line

Answer 3

$ seq 10 | tail -n +5 | head -n -2
5
6
7
8
$ seq 10 | awk '{p3=p2; p2=p1; p1=$0} NR>6{print p3}'
5
6
7
8
$ seq 10 | awk '{p[NR%6]=$0} NR>6{print p[(NR-2)%6]}'
5
6
7
8
$ seq 10 | awk -v b=4 -v a=2 'BEGIN{t=b+a} {p[NR%t]=$0} NR>t{print p[(NR-a)%t]}'
5
6
7
8
$ seq 10 | awk -v b=3 -v a=5 'BEGIN{t=b+a} {p[NR%t]=$0} NR>t{print p[(NR-a)%t]}'
4
5