当您在PHPMyAdmin中重命名数据库时，它会创建一个转储，然后删除并使用新名称重新创建数据库。

在phpmyadmin中，您可以轻松地重命名数据库

select database 

  goto operations tab

  in that rename Database to :

  type your new database name and click go

要求删除旧表和重新加载表数据单击确定

数据库被重命名

三个选项:

创建新数据库，关闭服务器，将文件从一个数据库文件夹移动到另一个数据库文件夹，然后重新启动服务器。注意，这只在所有表都是MyISAM的情况下才有效。 创建新的数据库，使用Create TABLE…LIKE语句，然后使用INSERT…从语句中选择*。 使用mysqldump并重新加载该文件。

这里的大多数答案都是错误的，原因有两个:

不能只使用RENAME TABLE，因为可能存在视图和触发器。如果有触发器，RENAME TABLE将失败 如果你想“快速”(如问题中要求的)重命名一个大数据库，你不能使用mysqldump

Percona有一篇关于如何做到这一点的博客文章: https://www.percona.com/blog/2013/12/24/renaming-database-schema-mysql/

以及由西蒙·R·琼斯(Simon R Jones)发布的脚本(制作的?)我修复了在脚本中发现的一个错误。你可以在这里看到:

https://gist.github.com/ryantm/76944318b0473ff25993ef2a7186213d

以下是它的副本:

#!/bin/bash
# Copyright 2013 Percona LLC and/or its affiliates
# @see https://www.percona.com/blog/2013/12/24/renaming-database-schema-mysql/
set -e
if [ -z "$3" ]; then
    echo "rename_db <server> <database> <new_database>"
    exit 1
fi
db_exists=`mysql -h $1 -e "show databases like '$3'" -sss`
if [ -n "$db_exists" ]; then
    echo "ERROR: New database already exists $3"
    exit 1
fi
TIMESTAMP=`date +%s`
character_set=`mysql -h $1 -e "SELECT default_character_set_name FROM information_schema.SCHEMATA WHERE schema_name = '$2'" -sss`
TABLES=`mysql -h $1 -e "select TABLE_NAME from information_schema.tables where table_schema='$2' and TABLE_TYPE='BASE TABLE'" -sss`
STATUS=$?
if [ "$STATUS" != 0 ] || [ -z "$TABLES" ]; then
    echo "Error retrieving tables from $2"
    exit 1
fi
echo "create database $3 DEFAULT CHARACTER SET $character_set"
mysql -h $1 -e "create database $3 DEFAULT CHARACTER SET $character_set"
TRIGGERS=`mysql -h $1 $2 -e "show triggers\G" | grep Trigger: | awk '{print $2}'`
VIEWS=`mysql -h $1 -e "select TABLE_NAME from information_schema.tables where table_schema='$2' and TABLE_TYPE='VIEW'" -sss`
if [ -n "$VIEWS" ]; then
    mysqldump -h $1 $2 $VIEWS > /tmp/${2}_views${TIMESTAMP}.dump
fi
mysqldump -h $1 $2 -d -t -R -E > /tmp/${2}_triggers${TIMESTAMP}.dump
for TRIGGER in $TRIGGERS; do
    echo "drop trigger $TRIGGER"
    mysql -h $1 $2 -e "drop trigger $TRIGGER"
done
for TABLE in $TABLES; do
    echo "rename table $2.$TABLE to $3.$TABLE"
    mysql -h $1 $2 -e "SET FOREIGN_KEY_CHECKS=0; rename table $2.$TABLE to $3.$TABLE"
done
if [ -n "$VIEWS" ]; then
    echo "loading views"
    mysql -h $1 $3 < /tmp/${2}_views${TIMESTAMP}.dump
fi
echo "loading triggers, routines and events"
mysql -h $1 $3 < /tmp/${2}_triggers${TIMESTAMP}.dump
TABLES=`mysql -h $1 -e "select TABLE_NAME from information_schema.tables where table_schema='$2' and TABLE_TYPE='BASE TABLE'" -sss`
if [ -z "$TABLES" ]; then
    echo "Dropping database $2"
    mysql -h $1 $2 -e "drop database $2"
fi
if [ `mysql -h $1 -e "select count(*) from mysql.columns_priv where db='$2'" -sss` -gt 0 ]; then
    COLUMNS_PRIV="    UPDATE mysql.columns_priv set db='$3' WHERE db='$2';"
fi
if [ `mysql -h $1 -e "select count(*) from mysql.procs_priv where db='$2'" -sss` -gt 0 ]; then
    PROCS_PRIV="    UPDATE mysql.procs_priv set db='$3' WHERE db='$2';"
fi
if [ `mysql -h $1 -e "select count(*) from mysql.tables_priv where db='$2'" -sss` -gt 0 ]; then
    TABLES_PRIV="    UPDATE mysql.tables_priv set db='$3' WHERE db='$2';"
fi
if [ `mysql -h $1 -e "select count(*) from mysql.db where db='$2'" -sss` -gt 0 ]; then
    DB_PRIV="    UPDATE mysql.db set db='$3' WHERE db='$2';"
fi
if [ -n "$COLUMNS_PRIV" ] || [ -n "$PROCS_PRIV" ] || [ -n "$TABLES_PRIV" ] || [ -n "$DB_PRIV" ]; then
    echo "IF YOU WANT TO RENAME the GRANTS YOU NEED TO RUN ALL OUTPUT BELOW:"
    if [ -n "$COLUMNS_PRIV" ]; then echo "$COLUMNS_PRIV"; fi
    if [ -n "$PROCS_PRIV" ]; then echo "$PROCS_PRIV"; fi
    if [ -n "$TABLES_PRIV" ]; then echo "$TABLES_PRIV"; fi
    if [ -n "$DB_PRIV" ]; then echo "$DB_PRIV"; fi
    echo "    flush privileges;"
fi

将它保存到一个名为rename_db的文件中，并使用chmod +x rename_db使脚本可执行，然后像。/rename_db localhost old_db new_db那样使用它

这里已经有很多非常好的答案，但我没有看到PHP版本。这将在大约一秒钟内复制一个800M DB。

$oldDbName = "oldDBName";
$newDbName = "newDBName";
$oldDB     = new mysqli("localhost", "user", "pass", $oldDbName);
if($oldDB->connect_errno){
    echo "Failed to connect to MySQL: (" . $oldDB->connect_errno . ") " . $oldDB->connect_error;
    exit;
}
$newDBQuery = "CREATE DATABASE IF NOT EXISTS {$newDbName}";
$oldDB->query($newDBQuery);
$newDB = new mysqli("localhost", "user", "pass");
if($newDB->connect_errno){
    echo "Failed to connect to MySQL: (" . $newDB->connect_errno . ") " . $newDB->connect_error;
    exit;
}

$tableQuery  = "SHOW TABLES";
$tableResult = $oldDB->query($tableQuery);
$renameQuery = "RENAME TABLE\n";
while($table = $tableResult->fetch_array()){
    $tableName = $table["Tables_in_{$oldDbName}"];
    $renameQuery .= "{$oldDbName}.{$tableName} TO {$newDbName}.{$tableName},";
}
$renameQuery = substr($renameQuery, 0, strlen($renameQuery) - 1);
$newDB->query($renameQuery);

I posted this How do I change the database name using MySQL? today after days of head scratching and hair pulling. The solution is quite simple export a schema to a .sql file and open the file and change the database/schema name in the sql CREAT TABLE section at the top. There are three instances or more and may not be at the top of the page if multible schemas are saved to the file. It is posible to edit the entire database this way but I expect that in large databases it could be quite a pain following all instances of a table property or index.