I have a function in bash that get param a string, for example:
MYSQL_DATABASE then I want in my file to create a var named VAR_MYSQL_DATABASE but I can`t get the value.
create_var()
{
read -p "Enter $1 : " VAR_$1
printf VAR_$1 // print VAR_MYSQL_DATABASE_NAME instead what I typed, so how do I get the value?
if [[ -z VAR_$1 ]]; then
printf '%s\n' "No input entered. Enter $1"
create_entry $1
fi
}
create_var "MYSQL_DATABASE_NAME"
Use the declare built-in in bash,
name="MYSQL_DATABASE"
declare VAR_${name}="Some junk string"
printf "%s\n" "${VAR_MYSQL_DATABASE}"
Some junk string
With the above logic, you can modify how your name variable is controlled, either present locally. If it is passed as argument from a function/command-line do
declare VAR_${1}="Your string value here"
Perhaps you want to achieve something like this,
create_var()
{
read -p "Enter value: " value
declare -g VAR_$1="$value"
dynamVar="VAR_$1"
if [[ -z "${!dynamVar}" ]]; then
printf '%s\n' "No input entered. Enter $1"
create_entry $1
fi
}
So, here we are creating the dynamic variable using the declare built-in and with the dynamic part VAR_$1 cannot be referenced directly as normal variables, hence using indirect expansion {!variable} to see if the created variable is available.
$1
declare makes the variable local to the function, unless you use the -g option. If it's local to the function, there's no real need for it to have any specific name.Try the following :
#!/bin/bash
create_var() {
declare -g VAR_$1
ref=VAR_$1
read -p "Enter $1: " "VAR_$1"
echo "inside : ${!ref}"
}
create_var "MYSQL_DATABASE_NAME"
echo "output : $VAR_MYSQL_DATABASE_NAME"
declare -g will make sure the variable exists outside of the function scope, and "VAR_$1" is used to dynamically create the variable names.
Output :
Enter MYSQL_DATABASE_NAME: Apple
inside : Apple
output : Apple
echo "inside : $VAR_MYSQL_DATABASE_NAME" is not dynamic. MYSQL_DATABASE_NAME is a param. So somehow I should use echo "inside : $VAR_$1"
declare -g would be preferable to export in versions of bash that support it.read -p "Enter $1: " "VAR_$1" is sufficient; there is no need for a command substitution here.You can use the printf function with the -v option to "print" to a variable name.
create_var()
{
while : ; do
IFS= read -r -p "Enter $1 : " value
if [[ -n $value ]]; then
printf -v "VAR_$1" '%s' "$value"
return
fi
printf 'No input entered. Enter %s\n' "$1"
done
}
(I've rewritten your function with a loop to avoid what appeared to be an attempt at recursion.)
MYSQL_DATABASE_NAMEand set a variable in the bash script with the name. The idea is that I have to ask 15 variables and because of this I made this function to be dry. If I use this lineread -p "Enter MYSQL_USER_NAME : " MYSQL_DATABASE_NAMEall works.