Dynamic variables with two variables

Question

I have a bash scripting question. I am trying to create the following code, however, it does not work properly. This leads me to wonder if it is even possible. I hope someone on here can figure it out.

MYSQL_DB_MyDatabase_USERS_username_PASSWORD=Password

DATABASE_NAME=MyDatabase
DATABASE_USER=username
DATABASE_PASS_TEMP=MYSQL_DB_${DATABASE_NAME}_USERS_${DATABASE_USER}_PASSWORD

echo $DATABASE_PASS_TEMP

I would want the output to be "Password" (the original variable). I am getting an error and I know I have incorrect syntax. I have googled around and found that using one variable in the variable name, one would use echo "${!DATABASE_PASS_TEMP}", but that does not work when using two dynamic variables. Thanks in advance for any help you can give.

anubhava · Accepted Answer · 2014-04-08 20:36:13Z

2

Use BASH's variable indirection:

echo "${!DATABASE_PASS_TEMP}"
Password

answered Apr 8, 2014 at 20:36

anubhava

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2 Comments

alvits Over a year ago

+1 Actually it is more known as variable indirection. But your answer is correct nonetheless.

anubhava Over a year ago

Glad to know, can you mark the answer as accepted by clicking on tick mark on top-left of my answer.

glenn jackman · Accepted Answer · 2014-04-08 20:55:40Z

0

Using an associative array would tidy things up some:

declare -A MYSQL_DB_PASSWORDS=(
    [MyDatabase,username]=Password
)

DATABASE_NAME=MyDatabase
DATABASE_USER=username

DATABASE_PASS_TEMP=${MYSQL_DB_PASSWORDS["$DATABASE_NAME","$DATABASE_USER"]}
echo "$DATABASE_PASS_TEMP"

Password

answered Apr 8, 2014 at 20:55

glenn jackman

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