Solved

use a variable (like $i) as part of another variables name in bash

Posted on 2013-06-20
5
238 Views
Last Modified: 2013-06-23
I have a file with variables like this...
server_1=
port_1=
user_1=

server_2=
port_2=
user_2=

Open in new window


I am trying to write a script that can read them in and manipulate them and write them back out.  And it should be able to handle any number of them.  Below is a simplified version of what I want to be able to do.  (i.e. not actually trying to edit anything here.  Just write it back out)

Is this at all possible?


source filewithvariables
for (( i=1; i<=$number_of_servers; i++ ))
do

echo "server_${i}=${server_${i}}" >> output 
echo "port_${i}=${port_${i}}" >> output 
echo "user_${i}=${user_${i}}" >> output 

done

Open in new window

0
Comment
Question by:Xetroximyn
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 3
  • 2
5 Comments
 
LVL 84

Accepted Solution

by:
ozo earned 500 total points
ID: 39263887
eval echo "server_${i}=\${server_${i}}" >> output
or
name=server_${i}; echo "server_${i}=${!name}" >> output
0
 
LVL 84

Expert Comment

by:ozo
ID: 39263920
or
for name in ${!server_*}  ; do  echo "$name=${!name}"  ; done
0
 

Author Comment

by:Xetroximyn
ID: 39266692
Thanks!

Just want to understand this a bit better... I think I understand the val method.  Baiscally eval replaces the ${i} with the "1" and leaves the first $ in there because it's escaped and thus bash is left to execute
server_1=${server_1}

And I think I understand the indirect referencing with the ${!name} syntax.  It seems like it takes the contents of $name and uses that for the variable name to get the contents of instead of just getting the contents of ${!name}

However - if it worked like I think it does - I don't quite understand why these two things would not be the same

echo $name_1
i=1
echo $name_${!i}

Open in new window


any chance you can help me understand that?
0
 
LVL 84

Expert Comment

by:ozo
ID: 39267231
i=1
echo $name_${!i}
would have the same effect as
echo $name_$1
0
 

Author Closing Comment

by:Xetroximyn
ID: 39269814
TY!
0

Featured Post

What does it mean to be "Always On"?

Is your cloud always on? With an Always On cloud you won't have to worry about downtime for maintenance or software application code updates, ensuring that your bottom line isn't affected.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Utilizing an array to gracefully append to a list of EmailAddresses
Originally, this post was published on Monitis Blog, you can check it here . In business circles, we sometimes hear that today is the “age of the customer.” And so it is. Thanks to the enormous advances over the past few years in consumer techno…
Get a first impression of how PRTG looks and learn how it works.   This video is a short introduction to PRTG, as an initial overview or as a quick start for new PRTG users.
In a recent question (https://www.experts-exchange.com/questions/29004105/Run-AutoHotkey-script-directly-from-Notepad.html) here at Experts Exchange, a member asked how to run an AutoHotkey script (.AHK) directly from Notepad++ (aka NPP). This video…

623 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question