Date: Sun, 17 Dec 2000 20:00:51 -0800 From: David <davidd@datasphereweb.com> To: Aaron Hill <hillaa@hotmail.com> Cc: questions@freebsd.org Subject: Re: Script: Variable substition within a variable? Message-ID: <20001217200051.A97413@datasphereweb.com> In-Reply-To: <F175g2UVe3lQN5vnoqt00000f22@hotmail.com>; from hillaa@hotmail.com on Mon, Dec 18, 2000 at 03:53:19AM %2B0000 References: <F175g2UVe3lQN5vnoqt00000f22@hotmail.com>
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On Mon, Dec 18, 2000 at 03:53:19AM +0000, Aaron Hill wrote: > Hello! > > Can anyone help with a script problem I'm having. Please let me explain what > I'm trying to do with two example files... > > fileone: > $name > > > filetwo: > #!/bin/sh > > name="Aaron" > fileone=`cat fileone` > > echo "Your name is $fileone" > > > When you execute filetwo this is the output you get: > Your name is $name > > The output I would like to get is: > Your name is Aaron > > > Can anyone point out if I've missed something fundamental or if there's > something that could help me do this substition. I think what you're looking for is something along the lines of: echo "Your name is " $fileone -- |> /\ \/ @ davidd@datasphereweb.com DataSphere - Back end web programming, site security, and networking david.daugherty@netmanage.com Software Engineer NetManage - The Bridge to E-Business http://www.wcug.wwu.edu/~doc ICQ: 21106703 "I like the dreams of the future better than the history of the past" -Thomas Jefferson To Unsubscribe: send mail to majordomo@FreeBSD.org with "unsubscribe freebsd-questions" in the body of the message
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