How to use an argument/parameter name as a variable in a bash script

Question

I'm trying to write a script that allows connection to various servers, e.g.

#!/bin/bash
# list of servers
server1=10.10.10.10
server2=20.20.20.20
ssh ${$1}

And I'd like to run it like:

sh connect.sh server1

Can't figure out how to use the parameter's name as a variable. Arrays do not work on my Ubuntu too.

score 8 · Accepted Answer · answered Oct 15 '13 at 10:16

8

Use shell indirection like this:

x=5
y=x
echo ${!y}
5

For your script, following works:

#!/bin/bash
# list of servers
server1=10.10.10.10
server2=20.20.20.20

arg1="$1"
ssh ${!arg1}

answered Oct 15 '13 at 10:16

anubhava

761,203
64
569
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1

+1 Really nice trick though it should be mentioned that this is bash only. – helpermethod Oct 15 '13 at 10:23
1

@helpermethod: Thanks and yes this is for BASH since question was tagged as BASH. – anubhava Oct 15 '13 at 10:29
Hmm getting error "Bad substitution" here ssh ${!arg1}. Any ideas? – Yasen Oct 15 '13 at 10:37
1

Don't run it `sh connect.sh server1` use it like `bash connect.sh server1` – anubhava Oct 15 '13 at 10:40
1

@Lamy It seems you're using `sh` and not `bash`. – devnull Oct 15 '13 at 10:40
1

+1, but if indirect parameter expansion is available, arrays should be as well. – chepner Oct 15 '13 at 12:24

helpermethod · Answer 2 · 2013-10-15T10:23:45.713

1

Easiest way would be to switch on $1:

case "$1" in
  server1) ssh "$server1"
           ;;
  server2) ssh "$server2"
           ;;
  *) ssh "$server1" # when no parameter is passed default to server1
     ;;
esac

edited Oct 15 '13 at 10:23

answered Oct 15 '13 at 10:16

helpermethod

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score 0 · Answer 3 · answered Oct 15 '13 at 10:22

0

Try this:

#!/bin/bash

# list of servers
server1=10.10.10.10
server2=20.20.20.20

if [ "$1" == "server1" ]; then 
    ssh $server1;
elif [ "$1" == "server2" ]; then 
    ssh $server2;
fi

answered Oct 15 '13 at 10:22

bbrady

1

How to use an argument/parameter name as a variable in a bash script

3 Answers3

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