CODEWITHSUNDEEP
stringbashvariablesconcatenation
Im Fine
Im Fine

Reputation: 273

Bash, Concatenating 2 strings to reference a 3rd variable

I have a bash script I am having some issues with concatenating 2 variables to call a 3rd.

Here is a simplification of the script, but the syntax is eluding me after reading the docs.

server_list_all="server1 server2 server3";
var1 = "server";
var2 = "all";

echo $(($var1_list_$var2));

This is about as close as I get to the right answer, it acknowledges the string and tosses an error on tokenization.

syntax error in expression (error token is "server1 server2 server3....

Not really seeing anything in the docs for this, but it should be doable.

EDIT: Cleaned up a bit

Upvotes: 27

Views: 27827

Answers (2)

Recurse
Recurse

Reputation: 3595

The Advanced Bash Scripting Guide has the answer for you (http://tldp.org/LDP/abs/html/ivr.html). You have two options, the first is classic shell:

 #!/bin/bash

 server_list_all="server1 server2 server3";
 var1="server";
 var2="all";

 server_var="${var1}_list_${var2}"
 eval servers=\$$server_var;

 echo $servers

Alternatively you can use the bash shortcut ${!var}

 #!/bin/bash

 server_list_all="server1 server2 server3";
 var1="server";
 var2="all";

 server_var="${var1}_list_${var2}"
 echo ${!server_var}

Either approach works.

Upvotes: 9

Eric Smith
Eric Smith

Reputation: 5392

The Bash Reference Manual explains how you can use a neat feature of parameter expansion to do some indirection. In your case, you're interested in finding the contents of a variable whose name is defined by two other variables:

server_list_all="server1 server2 server3"
var1=server
var2=all
combined=${var1}_list_${var2}

echo ${!combined}

The exclamation mark when referring to combined means "use the variable whose name is defined by the contents of combined"

Upvotes: 45

Related Questions