remove blank lines in awk display

I was trying to figure out a solution for this question. I wanted to use awk for the solution.

My input file is something like below.

-bash-3.2$ cat file
ramesh
ramesh_venkat
ramesh3_venkat3
ramesh4
ramesh5
venkat
venkat3
venkat4

I used awk command to extract the second values after _ as below.

awk -F "_" '{print $2}' file

However, though the above command prints the correct values I am getting blank lines in my output. I have 2 questions.

Question 1

How can I remove the blank lines in output so that I get only venkat and venkat3 in the output?

If I use printf instead of print in my awk, I get venkatvenkat3 as output which is not I wanted to achieve. I want the output like,

venkat
venkat3

Question 2

Using those values as an associative array or something, how can I find if the values actually occur in $1 column?

I wanted to achieve something like,

awk -F "_" '$2==1{print $1}' file

EDIT

I did not notice the awk solution of Stephane. Is it doing the same thing that I had mentioned?

Answers:

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Method 1

Another approach:

Question 1

awk -F_ '$2{print $2}' file

This will only print if $2 is defined. It is a shorter way of writing:

awk -F_ '{if($2){print $2}}' file

Question 2

Don’t have anything to add that has not already been addressed.

Method 2

Question1

$ awk -F _ 'NF > 1 {print $2}' file
venkat
venkat3

Question2

$ awk -F _ '
    NR == FNR {a[$1];next}
    ($2 in a) {print $2}
' file file
venkat
venkat3

Method 3

for Question 1, you could use the --only-delimited (-s) option of cut

cut -s -f2 -d'_' file
venkat
venkat3

Method 4

question 1

awk -F "_" '/_/ {print $2}' file

question 2

awk -F "_" '{values[$1]=1;}; END {for (val in values) print val;}' file

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0