mjhjmtu

I'm trying with the below code to print Available if there's a match, else nil

grep -o 'pattern' test.log | awk 'if($0=="pattern") print "Available"; else print "nil"'

The if part is working fine, but I'm not getting the else part if the grepped out is null.

If the pattern did not match, then grep produces no output and the awk program has no data to work with. This is why you will never get nil from the awk code.

Another way of doing this would be

if grep -q 'pattern' test.log; then
 echo 'Available'
else
 echo 'nil'
fi

The -q option of grep is used to stop the utility from generating any output (apart from possibly diagnostic output). Here, we don't want output from grep, only its exit status.

With awk, you could still do your test if you wish, but you will have to output the nil string conditionally in an END block:

grep -o 'pattern' test.log |
awk '/pattern/ print "Available"; found = 1 
 END if (!found) print "nil" '

The END block will be executed even if the awk script does not get any input to work with.

In fact, you could do the whole thing with awk:

awk '/pattern/ print "Available"; found = 1; exit 
 END if (!found) print "nil" ' test.log

Calling exit will invoke the END block, so we can't get rid of the found variable.