Check if mysql database exists, perform action based on result

Question

Is it possible from within a bash script to check if a mysql database exists. Depending on the result then perform another action or terminate the script?

Answer 1

It's easy enough to reliably tell if the database exists with mysqlshow. The trick is being able to reliably tell the difference between a database not existing, or some other failure. The version of mysqlshow I have exits with a '1' in either case, so it can't tell.

Here's what I came up with to handle it. Adjust your mysqlshow command accordingly, or put your credentials in to a chmod 600'd ~/.my.cnf file.

This works on Ubuntu 12 + 14. I haven't tested it in other environments yet:

#!/bin/bash -u

# Takes 1 argument. Aborts the script if there's a false negative.
function mysql_db_exists () {
  local DBNAME="$1"
  # Underscores are treated as wildcards by mysqlshow.
  # Replace them with '\\_'. One of the underscores is consumed by the shell to keep the one mysqlshow needs in tact.
  ESCAPED_DB_NAME="${DBNAME//_/\\\_}"
  RESULT="$(mysqlshow "$ESCAPED_DB_NAME" 2>&1)"; EXITCODE=$?
  if [ "$EXITCODE" -eq 0 ]; then
    # This is never a false positive.
    true
  else
    if echo "$RESULT" | grep -iq "Unknown database"; then
      # True negative.
      false
    else
      # False negative: Spit out the error and abort the script.
      >&2 echo "ERR (mysql_db_exists): $RESULT"
      exit 1
    fi
  fi
}

if mysql_db_exists "$1"; then
  echo "It definitely exists."
else
  echo "The only time you see this is when it positively does not."
fi