How to move grep inside awk script? - awk

Question

In the below have I 3 grep commands that I would like to replace with awk's grep. so I have tried
! /000000000000/;
! /000000000000/ $0;
! /000000000000/ $3;
where I don't get an error, but testing with both the script below and
$ echo 000000000000 | awk '{ ! /000000000000/; print }'
000000000000
it doesn't skip the lines as expected.
Question
Can anyone explain why my "not grep" doesn't work in awk?
grep -v '^#' $hosts | grep -E '[0-9A-F]{12}\b' | grep -v 000000000000 | awk '{
print "host "$5" {"
print " option host-name \""$5"\";"
gsub(/..\B/,"&:", $3)
print " hardware ethernet "$3";"
print " fixed-address "$1";"
print "}"
print ""
}' > /etc/dhcp/reservations.conf

Answer 1

Could you please try changing your code to:
echo 000000000000 | awk '!/000000000000/'
Problem in your attempt: $ echo 000000000000 | awk '{ ! /000000000000/; print }' Since you are checking condition ! /000000000000/ which is having ; after it so that condition works well and DO NOT print anything. But then you have print after it which is NOT COMING under that condition so it simply prints that line.
awk works on pattern{action} if you are putting semi colon in between it means that condition ends before it and statement after ; is all together a new statements for awk.
EDIT: Adding possible solution by seeing OP's attempt here, not tested at all since no samples are shown by OP. Also I am using --re-interval since my awk version is old you could remove in case you have new version of awk in your box.
awk --re-interval '!/^#/ && !/000000000000/ && /[0-9A-Fa-f]{12}/{
print "host "$5" {"
print " option host-name \""$5"\";"
gsub(/..\B/,"&:", $3)
print " hardware ethernet "$3";"
print " fixed-address "$1";"
print "}"
print ""
}' "$host" > /etc/dhcp/reservations.conf

Answer 2

Taking a look at your code:
$ echo 000000000000 | awk '
{
! /000000000000/ # on given input this evaluates to false
# but since its in action, affects nothing
print # this prints the record regardless of whatever happened above
}'
Adding a print may help you understand:
$ echo 000000000000 | awk '{ print ! /000000000000/; print }'
0
000000000000
Removing the !:
$ echo 000000000000 | awk '{ print /000000000000/; print }'
1
000000000000
This is all I can help you with since there is not enough information for more.