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From: ciemo@bananapc.wpd.sgi.com (Dave Ciemiewicz)
Newsgroups: comp.sys.sgi
Subject: Re: awk question...
Message-ID: <8792@odin.corp.sgi.com>
Date: 11 Jun 90 16:32:07 GMT
References: <9006091601.aa04003@VGR.BRL.MIL>
Sender: news@odin.corp.sgi.com
Reply-To: ciemo@bananapc.wpd.sgi.com (Dave Ciemiewicz)
Organization: Silicon Graphics, Inc.
Lines: 44

In article <9006091601.aa04003@VGR.BRL.MIL>, Claude.P.Cantin@NRC.CA writes:
> 
> I'm writting a script in which a variable takes the value of a userid.
> I then want to find out who this userid refers to.
> 
> I want to do that in one line, involving awk (I know how to do it using
> multiple lines of code).
> 
> If the userid is 123, the following would do just fine:
> 
>    awk -F: '$3 == 123 {print $1}' /etc/passwd
> 
> BUT 123 is the content of a variable, say UID.  The following does NOT
> work:
> 
>    awk -F: '$3 == $UID {print $1}' /etc/passwd
> 
> (the output is NOTHING).
> 
> I have tried several variations, including "$UID", and "$3"=="$UID", etc.,
> but none worked...
> 
> Anyone has an insight????
> 
> Thank you,
> 
>        Claude Cantin (CANTIN@VM.NRC.CA)

Claude,

Your problem is that shell variables are not expanded within single quotes (').

Try the following:

    awk -F: '$3 == '$UID' {print $1}' /etc/passwd

You can also use:

    awk -F: '$3 == uid {print $1}' uid=$UID /etc/passwd

You might check out the AWK book by Aho, Kernighan, and Weinberger from
Addison-Wesley for more fun with AWK.

						--- Ciemo