Qusetion about awk

[blues77]

Hello,

I'm trying to use awk to parse the uid=xxx information that is displayed when I run the command id $user. Right now my code looks like

#!/bin/bash
                                                                                                                                                                
id $USER | awk '/uid=[0-9]*/ {print $1}'

Which prints out:
uid=500(jsmith)


My question is how do I get awk to just print out the uid=xxx portion of the 'id $USER' command. In other words to leave off the jsmith portion. Any help is greatly
appreciated.

 

[Ygor]

Perhaps you can use: [tt] id -u $USER[/tt]

 

[blues77]

Thanks for the tip Ygor. It worked. Now I need a way of displaying the "real name" of the user. I thought about using the finger command with something like

finger | awk '// {print $1}'

However this will print the name of the user several times if they have more than one terminal open. Can anyone suggest another way for me to accomplish this. Any help is always appreciated.

Thanks!

 

[futurelet]

[tt]
finger | awk '{a[$1]++}END{for(k in a)print k}'
[/tt]

 

[Ygor]

I think finger just does a look-up to /etc/passwd - you could do something similar...
[tt]
USERID=$(id -u $USER)
USERNAME=$(awk -F: "$USERID"'==$3{print $5}' /etc/passwd)
[/tt]

 

[PHV]

What about simply this ?
awk -F: '$1=="'"$USER"'"{printf "%s (uid=%d,gid=%d) %s\n",$1,$3,$4,$5}' /etc/passwd

Hope This Helps, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884 or FAQ222-2244

 

[kHz]

A couple of ways:

id $USER | exec 3<&0
while read -u3 line; do
   print ${line%%\(*}
done

id $USER | awk -F'(' '{print $1}'