awk for multiple patterns

Question

My file looks like:

L   0   256 *   *   *   *   *
H   0   307 100.0   +   0   0
S   30  351 *   *   *   *   *
D   8   27  *   *   *   *   99.3    
C   11  1   *   *   *   *   *   

for my script I would like to start by awk print $0 for certain lines using $1

Such as

awk '{if ($1!="C") {print $0}  else if ($1!="D") {print $0}}'

But, there has to be a way to combine "C" and "D" into one IF statement... right?

For example if I want to search for == L,H,S ie... NOT C or D how would I right this?

Answer 1

Your present condition is not correct as both $1!="C" and $1!="D" can't be false at the same time. Hence, it will always print the whole file.

This will do as you described:

awk '{if ($1!="C" && $1!="D") {print $0}}'  file

Answer 2

Using awk, you can provide rules for specific patterns with the syntax

awk 'pattern {action}' file

see the awk manual page for the definition of a pattern. In your case, you could use a regular expression as a pattern with the syntax

awk'/regular expression/ {action}' file

and a basic regular expression which would suit your needs could be

awk '/^[^CD]/ {print $0}' file

which you can actually shorten into

awk '/^[^CD]/' file

since {print $0} is the default action, as suggested in the comments.

Answer 3

awk '$1 ~ /[CD]/' file

awk '$1 ~ /[LHS]/' file

awk '$1 ~ /[^LHS]/' file

awk '$1 !~ /[LHS]/' file