awk printing against the condition - awk

Question

I have a simple question but I could not figure it out.
I have a file that I want to print all the lines that DO NOT match the condition I specify in the awk if condition. But I can just get to print the condition, how the other would work?
This is my code:
awk '{if ($18==0 && $19==0 && $20==0 && $21==0) print $0}' file
I also tried this:
awk '{if !($18==0 && $19==0 && $20==0 && $21==0) print $0}' file
But the second one doesn't work, any help is appreciated. Thank you.

Answer 1

Here you can do:
awk '$18+$19+$20+$21!=0' file
print $0 is not needed, since its default action.

Answer 2

The negation (!) needs to be inside the parentheses:
awk '{if (!($18==0 && $19==0 && $20==0 && $21==0)) print $0}' file
And we add another set inside to wrap everything.
(FYI, if you had given how it "didn't work" (i.e., a syntax error on !, that would have been more helpful. Please remember to include error messages or symptoms of something not working for future questions!)

Answer 3

You could also reverse your conditional statement:
you want the opposite of :
awk '{if ($18==0 && $19==0 && $20==0 && $21==0) print $0}' file
Which can either be :
awk '{if ($18!=0 || $19!=0 || $20!=0 || $21!=0) print $0}' file
or
awk '{if (!($18==0 && $19==0 && $20==0 && $21==0)) print $0}' file
Example :
!cat file
A 0 0 0
B 1 1 1
C 1 0 1
awk '$2+$3+$4!=0' file
B 1 1 1
C 1 0 1
awk '{if ($2!=0 || $3!=0 || $4!=0) print $0}' file
B 1 1 1
C 1 0 1
awk '{if (!($2==0 && $3==0 && $4==0)) print $0}' file
B 1 1 1
C 1 0 1
awk '{if (!($2==0 || $3==0 || $4==0)) print $0}' file
B 1 1 1