How to check if mysql database exists

SELECT SCHEMA_NAME
  FROM INFORMATION_SCHEMA.SCHEMATA
 WHERE SCHEMA_NAME = 'DBName'

If you just need to know if a db exists so you won't get an error when you try to create it, simply use (From here):

CREATE DATABASE IF NOT EXISTS DBName;

From the shell like bash

if [[ ! -z "`mysql -qfsBe "SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME='db'" 2>&1`" ]];
then
  echo "DATABASE ALREADY EXISTS"
else
  echo "DATABASE DOES NOT EXIST"
fi

If you are looking for a php script see below.

$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
  die('Not connected : ' . mysql_error());
}

// make foo the current db
$db_selected = mysql_select_db('foo', $link);
if (!$db_selected) {
  die ('Cannot use foo : ' . mysql_error());
}

With this Script you can get Yes or No database exists, in case it does not exist it does not throw Exception.

SELECT 
    IF(EXISTS( SELECT 
                SCHEMA_NAME
            FROM
                INFORMATION_SCHEMA.SCHEMATA
            WHERE
                SCHEMA_NAME = 'DbName'),
        'Yes',
        'No')  as exist

Here's my way of doing it inside a bash script:

#!/bin/sh

DATABASE_USER=*****
DATABASE_PWD=*****
DATABASE_NAME=my_database

if mysql -u$DATABASE_USER -p$DATABASE_PWD -e "use $DATABASE_NAME";
then
echo "Database $DATABASE_NAME already exists. Exiting."
exit
else
echo Create database
mysql -u$DATABASE_USER -p$DATABASE_PWD -e "CREATE DATABASE $DATABASE_NAME"
fi

Be careful when checking for existence with a like statement!

If in a series of unfortunate events your variable ends up being empty, and you end up executing this:

SHOW DATABASES like '' -- dangerous!

It will return ALL databases, thus telling the calling script that it exists since some rows were returned.

It's much safer and better practice to use an "=" equal sign to test for existence.

The correct and safe way to test for existence should be:

SHOW DATABASES WHERE `database` = 'xxxxx' -- safe way to test for existence

Note that you have to wrap the column name database with backticks, it can't use relaxed syntax in this case.

This way, if the code creating the variable 'xxxxx' returned blank, then SHOW DATABASES will not return ALL databases, but will return an empty set.

Another php solution, but with PDO:

<?php
try {
   $pdo = new PDO('mysql:host=localhost;dbname=dbname', 'root', 'password', [ PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION ]);
   echo 'table dbname exists...';
}
catch (PDOException $e) {
   die('dbname not found...');
}

SELECT COUNT(*) FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = 'DbName'

1 - exists, 0 - not

Golang solution

create a test package and add:

import "database/sql"

// testing database creation
func TestCreate(t *testing.T){
    Createdb("*Testdb") // This just calls the **sql.DB obect *Testdb 
    db,err := sql.Open("mysql", "root:root@tcp(127.0.0.1:3306)/*Testdb")
    if err != nil{
        panic(err)
    }
    defer db.Close()
    _, err = db.Exec("USE *Testdb")
    if err != nil{
        t.Error("Database not Created")
    }

} 

Using the INFORMATION_SCHEMA or show databases is not reliable when you do not have enough permissions to see the database. It will seem that the DB does not exist when you just don't have access to it. The creation would then fail afterwards. Another way to have a more precise check is to use the output of the use command, even though I do not know how solid this approach could be (text output change in future versions / other languages...) so be warned.

CHECK=$(mysql -sNe "use DB_NAME" 2>&1)
if [ $? -eq 0 ]; then
  # database exists and is accessible
elif [ ! -z "$(echo $CHECK | grep 'Unknown database')" ]; then
  # database does not exist
elif [ ! -z "$(echo $CHECK | grep 'Access denied')" ]; then
  # cannot tell if database exists (not enough permissions)"
else
  # unexpected output
fi