piping search pattern to awk 1

[gibson153]

I want to see who is online and then print information about each user from /etc/passwd/

awk -f $(who -qn 1) '{ print $0 }' /etc/passwd

any suggestion on how I can make this work?

David

 

[CaKiwi]

This seems to work

who -qn 1 | awk '{print "/" $1 "/{print}"}' | awk -f - /etc/passwd

CaKiwi

 

[CaKiwi]

Upon further review, it would be better to search for the user only at the beginning of the line in passwd to avoid finding lines which had the same string as a user name further along. Also the {print} is not necessary as it is the default. This results in the following solution:

who -qn 1 | awk '{print "/^" $1 "/"}' | awk -f - /etc/passwd

I believe some older awks do not allow the program to come from stdin. In this case, you would need to redirect the output from the first awk to a temporary file and use this file as the program for the second awk.

 

[marsd]

Here you go:

awk ' BEGIN {
"who" | getline info
gsub(/tty.*/, "" , info)

while ((getline pots < &quot;/etc/passwd&quot > 0) {

split(pots, names, &quot;:&quot
for (x in names)
if (info ~ names[x]) {
print pots
}

}
}'

Needs to be filtered more, maybe you can figure it out.
It does come up with the right info+.
gawk 3.1.0

 

[gibson153]

Thanks this worked perfect!!

David