How to group lines and show every nth line of that group

This is the file:

-
line1
line2
line3
line4
line5
line6
line7
-
line8
line9
line10
line11
line12
line13
line14
-
line15
line16
line17
line18
line19
line20
line21

This is what I tried so far:

awk 'NR%6==1' file

It shows every 6th line:

-
line6
line11
line16

But I expect this:

line6
line12
line18

The file lines are grouped by - each 7th line as the example.

you can solve it like this :

awk '/^-/{group++;next} group && ((NR-group)%6==0){print}' file

• '/^-/{group++;next}' - When a line starts with "-", increment the group counter and skip to the next line.
• 'group && ((NR-group)%6==0){print}' - If we are in a group (i.e., group is not 0) and the current line number minus the group number modulo 6 is 0, print the current line

Result :

line6
line12
line18

If you really wanted to print the nth line of every group then it'd be:

$ awk -v n=6 '/^-/{c=0} c++ == n' file
line6
line13
line20

$ awk -v n=7 '/^-/{c=0} c++ == n' file
line7
line14
line21

For every line which is not the separator, increment a counter. If the incremented counter is divisible by 6, print the line:

awk '$0 != "-" && ++c % 6 == 0'