Bash script take in parameter or present a select - bash

Question

We have a script we use to promote databases from dev to QA. Currently the script has a select statement we use to print out all the available databases and then takes your selection and does the various bits to promote the db.
I want to add a parameter to bypass that select statement for scripting purposes.
Here's how the script works currently:
# GET LIST OF DATABASES
DATABASES=`echo "show databases;"|/usr/bin/mysql -u root -h $DEVHOST|grep -v "information_schema"|grep -v "Database"|grep -v "performance_schema"|grep -v "mysql"`
eval set $DATABASES
select DEVDB in "$#"
do
if [[ "$DEVDB" =~ "dev1" ]]; then
QADB=`echo $DEVDB|sed -e s/_dev1_/_qa1_/`
fi
# Does various bits after this
Here's what I'm trying to do:
arg1=$1
if [ ! -z "$arg1" ]; then
DEVDB=$arg1
else
# GET LIST OF DATABASES
DATABASES=`echo "show databases;"|/usr/bin/mysql -u root -h $DEVHOST|grep -v "information_schema"|grep -v "Database"|grep -v "performance_schema"|grep -v "mysql"`
eval set $DATABASES
select DEVDB in "$#"
do
fi
if [[ "$DEVDB" =~ "dev1" ]]; then
QADB=`echo $DEVDB|sed -e s/_dev1_/_qa1_/`
fi
I'm getting an error at fi, I assume it's related to the select... do. How do I wrap these up in my if statements?

Answer 1

You need to do something in the body of the select. In this case it's sufficient to check that the variable has a value:
PS3="Select a database: "
select DEVDB; do
[[ $DEVDB ]] && break
done
echo "you selected $DEVDB"
Definitely don't need eval here -- it's always a code smell.
set -- $(
/usr/bin/mysql -u root -h $DEVHOST <<< "show databases;" |
grep -v -e information_schema -e Database -e performance_schema -e mysql
)