Removing trailing comma

[hannaman]

I have a command that gets information on a user account from a database and displays it similar to this format:


U A O Username Profilename Some other text
_ _ _ ________ ___________ _______________
U A O user profile_1 junk.junk.junk
U A O user profile_2 I don't care
Returned number something

I want to feed the profile names to another command that takes a comma separated list of profiles. I am able to get the list, but awk statement:

command | awk '$4 ~ /user/ { print $5 }{ OFS="," }'
I get the list with a trailing comma like:

profile_1,profile_2,
I have been able to remove the trailing comma with sed using:

command | awk '$4 ~ /user/ { print $5 }{ OFS="," }' | sed 's/,*$//'
Even though this works, sed complains of a "missing newline at end of standard input".
I have tried to remove the trailing comma with awk using either sub or gensub, but I have been unsuccessful.
Does anyone know of a cleaner way of getting a comma separated list without the trailing comma like this?

Thanks

[hannaman]

OK, I got this work but it is a little ugly:

command | awk '$4 ~ /user/ { print $5 }{ OFS="," }' | awk 'sub(/,+$/,"")'

I would like to do this with a single awk statement, but it works and doesn't spit out any errors.

[trent.josephsen]

OK, I got this work but it is a little ugly:

command | awk '$4 ~ /user/ { print $5 }{ OFS="," }' | awk 'sub(/,+$/,"")'

I would like to do this with a single awk statement, but it works and doesn't spit out any errors.

In Perl you could do

command | perl -nae 'push @ary, $F[4] if $F[3] =~ /user/; END { print join(",", @ary), $/ }'

Perhaps an Awk guru can suggest how to achieve the same in that language (probably more concisely).

[Compyx]

Unfortunately awk doesn't have a join() function, although it has a split() function.

I'd do something like this, similar to the perl version: I store the profiles in an array and then print all elements, except the last, with a comma attached and then print the last element without the comma attached (simulating a join() function):


command | awk \
'$4 ~ /user/ { p[i++] = $5 } \
END { for (k = 0; k < i - 1; k++) { printf "%s,", p[k] } print p[k] }'


It's ugly, but it's a single awk invocation ;)

[geirha]

You are currently using a regex to match the username, I'm guessing you really want to do a string comparison instead. $4 ~ /user/ will match "user", "userA" and "superuser" etc. If you compare it as two strings instead, it will only match "user":
awk '$4 == "user" { ... }'

[DaithiF]

no neater than the preceding solutions, but for variety:

command | awk '$4 == "user" { print $5}' | sed -e :a -e 'N;s/\n/,/;ta'

[hannaman]

You are currently using a regex to match the username, I'm guessing you really want to do a string comparison instead. $4 ~ /user/ will match "user", "userA" and "superuser" etc. If you compare it as two strings instead, it will only match "user":
awk '$4 == "user" { ... }'

Thanks for that information. Since the command that I run only shows information for the user that I specify, I am only matching the username to exclude header and footer data.