bash - How to check if a variable has "mysqli" or "mariadb" value -

i have following piece of script:

set ${db_type:='mysqli'}  echo $db_type  if [  $db_type -eq "mysqli"  -o  $db_type -eq "mariadb"  ];       #do stuff else   echo >&2 "this database type not supported"    echo >&2 "did forget -e db_type='mysqli' ^or^ -e db_type='mariadb' ?"   exit 1 fi 

but piece of script somehow fails on:

if [  $db_type -eq "mysqli"  -o  $db_type -eq "mariadb"  ]; 

so how can compare if $db_type id either "mysqli" or "mariadb"?

it useful know how fails there few things wrong following:

if [  $db_type -eq "mysqli"  -o  $db_type -eq "mariadb"  ]; 

• variables unquoted (dangerous within [)
• -eq comparing integers, = should used strings (see this question detailed discussion of difference)

as you've tagged bash should aware of improved [[ allows write this:

if [[ $db_type = mysqli || $db_type = mariadb ]]; 

within [[ don't need quote variables , can use || logical or.

you might consider using this:

case $db_type in     mysqli|mariadb)         # 1 thing         ;;     *)         # other stuff         ;; esac