who - uniq output

lochraven · #1 (**permalink**) 10-24-2008

Hi,
I'd like to have a script what takes the 'who' output and grabs the user names and outputs just the user name, and no duplicates.

I know I could do something like:

who | awk '{print $1}' | uniq -u

but I'd like to stay away from using the 'uniq' comand and just use awk.

Thanks

joeyg · #2 (**permalink**) 10-24-2008

The following works, but not sure if all necessary.

Code:

#! /usr/bin/bash

who | awk '{line[$1] = $1
     sum[$1]++}
    END{
     for(i in sum)
     if (sum[i]>0)
       print line[i]
}'

set array variable to the first read (which is name) field
increment a counter so I can know if anything in array value

step thru all values in my sum array
if the value in sum > 0 (meaning I have a match)
print out the corresponding entry in the line array

Franklin52 · #3 (**permalink**) 10-24-2008

Or:

Code:

who | awk '!($1 in a){a[$1];print $1}'

Regards

lochraven · #4 (**permalink**) 10-24-2008

thats awesome, thanks to the both of you

How about excluding 'root' from the output?

Franklin52 · #5 (**permalink**) 10-24-2008

Code:

who | awk '!($1 in a) && !/root/{a[$1];print $1}'

joeyg · #6 (**permalink**) 10-24-2008

Excluding root

Code:

> cat uniq_names
#! /usr/bin/bash

who | awk '$1!~"root" {line[$1] = $1
     sum[$1]++}
    END{
     for(i in sum)
     if (sum[i]>0)
       print line[i]
}'

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