Is it possible to check if a (MySQL) database exists after having made a connection.
I know how to check if a table exists in a DB, but I need to check if the DB exists. If not I have to call another piece of code to create it and populate it.
I know this all sounds somewhat inelegant - this is a quick and dirty app.
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;
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());
}
Here is a bash function for checking if a database exists:
function does_db_exist {
local db="${1}"
local output=$(mysql -s -N -e "SELECT schema_name FROM information_schema.schemata WHERE schema_name = '${db}'" information_schema)
if [[ -z "${output}" ]]; then
return 1 # does not exist
else
return 0 # exists
fi
}
Another alternative is to just try to use the database. Note that this checks permission as well:
if mysql "${db}" >/dev/null 2>&1 </dev/null
then
echo "${db} exists (and I have permission to access it)"
else
echo "${db} does not exist (or I do not have permission to access it)"
fi
For those who use php with mysqli then this is my solution. I know the answer has already been answered, but I thought it would be helpful to have the answer as a mysqli prepared statement too.
$db = new mysqli('localhost',username,password);
$database="somedatabase";
$query="SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME=?";
$stmt = $db->prepare($query);
$stmt->bind_param('s',$database);
$stmt->execute();
$stmt->bind_result($data);
if($stmt->fetch())
{
echo "Database exists.";
}
else
{
echo"Database does not exist!!!";
}
$stmt->close();
Long winded and convoluted (but bear with me!), here is a class system I made to check if a DB exists and also to create the tables required:
<?php
class Table
{
public static function Script()
{
return "
CREATE TABLE IF NOT EXISTS `users` ( `id` INT NOT NULL PRIMARY KEY AUTO_INCREMENT );
";
}
}
class Install
{
#region Private constructor
private static $link;
private function __construct()
{
static::$link = new mysqli();
static::$link->real_connect("localhost", "username", "password");
}
#endregion
#region Instantiator
private static $instance;
public static function Instance()
{
static::$instance = (null === static::$instance ? new self() : static::$instance);
return static::$instance;
}
#endregion
#region Start Install
private static $installed;
public function Start()
{
var_dump(static::$installed);
if (!static::$installed)
{
if (!static::$link->select_db("en"))
{
static::$link->query("CREATE DATABASE `en`;")? $die = false: $die = true;
if ($die)
return false;
static::$link->select_db("en");
}
else
{
static::$link->select_db("en");
}
return static::$installed = static::DatabaseMade();
}
else
{
return static::$installed;
}
}
#endregion
#region Table creator
private static function CreateTables()
{
$tablescript = Table::Script();
return static::$link->multi_query($tablescript) ? true : false;
}
#endregion
private static function DatabaseMade()
{
$created = static::CreateTables();
if ($created)
{
static::$installed = true;
}
else
{
static::$installed = false;
}
return $created;
}
}
In this you can replace the database name en with any database name you like and also change the creator script to anything at all and (hopefully!) it won't break it. If anyone can improve this, let me know!
Note
If you don't use Visual Studio with PHP tools, don't worry about the regions, they are they for code folding :P
Here's my way of doing it inside a bash script:
#!/bin/sh
DATABASE_USER=*****
DATABASE_PWD=*****
DATABASE_NAME=my_database
if mysql -u$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
Rails Code:
ruby-1.9.2-p290 :099 > ActiveRecord::Base.connection.execute("USE INFORMATION_SCHEMA")
ruby-1.9.2-p290 :099 > ActiveRecord::Base.connection.execute("SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = 'entos_development'").to_a
SQL (0.2ms) SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = 'entos_development'
=> [["entos_development"]]
ruby-1.9.2-p290 :100 > ActiveRecord::Base.connection.execute("SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = 'entos_development1'").to_a
SQL (0.3ms) SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = 'entos_development1'
=> []
=> entos_development exist , entos_development1 not exist
IF EXISTS (SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = N'YourDatabaseName')
BEGIN
-- Database exists, so do your stuff here.
END
If you are using MSSQL instead of MySQL, see this answer from a similar thread.
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.
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
