冗长而令人费解(但请容忍我!)，这里是一个类系统，我做的检查是否DB存在，并创建所需的表:

<?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;
    }
}

在这种情况下，您可以将数据库名称en替换为您喜欢的任何数据库名称，也可以将创建者脚本更改为任何名称，并且(希望!)不会破坏它。如果有人可以改进这一点，请告诉我!

请注意 如果你不使用Visual Studio和PHP工具，不要担心这些区域，它们是用于代码折叠的

有了这个脚本，你可以得到是否数据库存在，如果它不存在，它不会抛出异常。

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

从壳像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

对于那些使用php与mysqli然后这是我的解决方案。我知道答案已经回答了，但我认为这将是有帮助的答案作为一个mysqli准备的声明。

$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();

这是我在bash脚本中做这件事的方式:

#!/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代码:

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存在，entos_development1不存在