In Bash scripting, how can I add a variable into a variable name?

Question

So take the following script as an example:

PROFILE1=profileName
SERVER1=serverName
PROFILE2=profile2Name
SERVER2=server2Name

VAR_$PROFILE1_$SERVER1=/home/$PROFILE1/$SERVER1/data
VAR_$PROFILE1_$SERVER2=/home/$PROFILE1/$SERVER2/data
VAR_$PROFILE2_$SERVER1=/home/$PROFILE2/$SERVER1/data

echo "$VAR_$PROFILE1_$SERVER1"

That is the basics of what I am attempting to accomplish. This may not be the best way to go about it, but for someone who is fairly new to bash scripting this was my first thought on how to do it.

I need to be able to only change one location in my script file and be able to add new "VAR_..." as needed for the system I am on.

Is this even possible? If not then if you understand what I am attempting to accomplish could you suggest an alternative?

So you want to, for example, generate the variable $VAR_profileName_serverName by replacing what $PROFILE1 and $SERVER1 contain? — fedorqui
– fedorqui, Commented Mar 19, 2014 at 15:19

glenn jackman · Accepted Answer · 2014-03-19 15:40:24Z

3

If you have bash version 4, use an associative array:

declare -A data
data=(
  [$PROFILE1,$SERVER1]=/home/$PROFILE1/$SERVER1/data
  [$PROFILE1,$SERVER2]=/home/$PROFILE1/$SERVER2/data
  [$PROFILE2,$SERVER1]=/home/$PROFILE2/$SERVER1/data
)
for key in "${!data[@]}"; do
    printf "%s\t%s\n" "$key" "${data[$key]}"
done
echo "${data[$PROFILE1,$SERVER2]}"

profileName,server2Name /home/profileName/server2Name/data
profileName,serverName  /home/profileName/serverName/data
profile2Name,serverName /home/profile2Name/serverName/data
/home/profileName/server2Name/data

answered Mar 19, 2014 at 15:40

glenn jackman

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2 Comments

Brandon Stout Over a year ago

I like this approach much better than the way I was attempting to do it. Although the other approach mentioned by fedorqui works as well. Thanks!

Brandon Stout Over a year ago

I also have attempted this approach in my scenario and it worked out beautifully!

Community · Accepted Answer · 2017-05-23 10:31:48Z

3

You can use ~~eval~~ declare for this (thanks Glenn Jackman for pointing in the good direction!):

$ declare VAR_${PROFILE1}_${SERVER1}="hello"
$ echo $VAR_profileName_serverName
hello

In your case:

declare VAR_${PROFILE1}_${SERVER1}=/home/$PROFILE1/$SERVER1/data
declare VAR_${PROFILE1}_${SERVER2}=/home/$PROFILE1/$SERVER2/data
declare VAR_${PROFILE2}_${SERVER1}=/home/$PROFILE2/$SERVER1/data

Note the use of VAR_${PROFILE1}_${SERVER1}. The curly brackets tell what's the exact name of the variable, because otherwise VAR_$PROFILE1_$SERVER1 would look for a variable named PROFILE1_, etc.

As echo VAR_${PROFILE1}_${SERVER1} returns VAR_profileName_serverName, then you can make use of the variables, using the following:

$ var="VAR_${PROFILE1}_${SERVER1}"
$ echo ${!var}
/home/profileName/serverName/data

Note that ${!var} is a variable indirection, that fetches the variable whose name is in the $var variable.

edited May 23, 2017 at 10:31

CommunityBot

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answered Mar 19, 2014 at 15:25

fedorqui

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6 Comments

Brandon Stout Over a year ago

Ok, so to reference the VAR later on would it be possible to call "echo $VAR_$PROFILE1_$SERVER1"? The main reason is so that I do not have to copy and paste "serverName" all over the place, and keep it in one location.

fedorqui Over a year ago

@BrandonStout see my updated answer to see how to fetch the variables.

glenn jackman Over a year ago

I would use declare not eval

glenn jackman Over a year ago

Also, var="VAR_${PROFILE1}_${SERVER1}" is sufficient. You don't need to spawn an echo.

Brandon Stout Over a year ago

Ok thanks @fedorqui, I like the method described by glenn in his post better than my original method. But I was able to get this way to work as well. Thank you very much!

|

Mike S. · Accepted Answer · 2014-03-19 15:24:50Z

0

You mean this?

ABC="ahoj\$UNSETVAR"
echo $ABC
UNSETVAR=now_it_is_set
eval "DEV=$ABC"
echo $DEV

Output will be:

ahoj$UNSETVAR
ahojnow_it_is_set

answered Mar 19, 2014 at 15:24

Mike S.

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