Print unique of who command

awksortwho

I want to show the online users using the who command but I want to have a unique output and not show any duplicates. I piped the output into awk but I am not very familiar with that, is this the right way and how to proceed?

Best Answer

It's easier to just do it with sort:

 who | sort -u -k1,1

The -u flag asks for "unique" lines (suppress duplicates). The key flag (-k) says to only consider the first word in each line for purposes of sorting and uniqueness.

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Using awk:

$ awk 'seen[$0]++; seen[$0] == 2' file
Happy sad
Happy sad
Happy sad
Happy sad
Happy sad
Mad happy
Mad happy

This uses the text of each line as the key into the associative array seen. The first seen[$0]++ will cause a line that has been seen before to be printed since the value associated with the line will be non-zero on the second and subsequent times the line is seen. The seen[$0] == 2 causes the line to be printed again if this is the second time the line has been seen (without this, you'll miss one occurrence of each duplicated line).

This is related to awk '!seen[$0]++' which is sometimes used to remove duplicates without sorting (see e.g. How does awk '!a[$0]++' work?).

To only get one copy of the duplicated lines:

awk 'seen[$0]++ == 1' file

or,

sort file | uniq -d

Best Answer

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