没有本地访问服务器,有没有办法复制/克隆一个MySQL数据库(有内容和没有内容)到另一个不使用mysqldump?
我目前使用的是MySQL 4.0。
没有本地访问服务器,有没有办法复制/克隆一个MySQL数据库(有内容和没有内容)到另一个不使用mysqldump?
我目前使用的是MySQL 4.0。
当前回答
Mysqldump是个不错的解决方案。复制数据库的最简单方法:
mysqldump -uusername -ppass dbname1 |MySQL -uusername -ppass dbname2
此外,您可以通过以下方式更改存储引擎:
mysqldump -uusername -ppass dbname1 |sed 's/InnoDB/RocksDB/' |MySQL -uusername -ppass dbname2
其他回答
请注意,mysql实用工具....上有一个mysqldbcopy命令作为添加的一部分 https://dev.mysql.com/doc/mysql-utilities/1.5/en/utils-task-clone-db.html
如果你使用的是Linux,你可以使用这个bash脚本: (它可能需要一些额外的代码清理,但它工作…它比mysqldump|mysql快得多)
#!/bin/bash
DBUSER=user
DBPASSWORD=pwd
DBSNAME=sourceDb
DBNAME=destinationDb
DBSERVER=db.example.com
fCreateTable=""
fInsertData=""
echo "Copying database ... (may take a while ...)"
DBCONN="-h ${DBSERVER} -u ${DBUSER} --password=${DBPASSWORD}"
echo "DROP DATABASE IF EXISTS ${DBNAME}" | mysql ${DBCONN}
echo "CREATE DATABASE ${DBNAME}" | mysql ${DBCONN}
for TABLE in `echo "SHOW TABLES" | mysql $DBCONN $DBSNAME | tail -n +2`; do
createTable=`echo "SHOW CREATE TABLE ${TABLE}"|mysql -B -r $DBCONN $DBSNAME|tail -n +2|cut -f 2-`
fCreateTable="${fCreateTable} ; ${createTable}"
insertData="INSERT INTO ${DBNAME}.${TABLE} SELECT * FROM ${DBSNAME}.${TABLE}"
fInsertData="${fInsertData} ; ${insertData}"
done;
echo "$fCreateTable ; $fInsertData" | mysql $DBCONN $DBNAME
一个SQL,显示SQL命令,需要运行复制一个数据库从一个数据库到另一个。对于每个表,都有创建表语句和插入语句。它假设两个数据库都在同一台服务器上:
select @fromdb:="crm";
select @todb:="crmen";
SET group_concat_max_len=100000000;
SELECT GROUP_CONCAT( concat("CREATE TABLE `",@todb,"`.`",table_name,"` LIKE `",@fromdb,"`.`",table_name,"`;\n",
"INSERT INTO `",@todb,"`.`",table_name,"` SELECT * FROM `",@fromdb,"`.`",table_name,"`;")
SEPARATOR '\n\n')
as sqlstatement
FROM information_schema.tables where table_schema=@fromdb and TABLE_TYPE='BASE TABLE';
在PHP中:
function cloneDatabase($dbName, $newDbName){
global $admin;
$db_check = @mysql_select_db ( $dbName );
$getTables = $admin->query("SHOW TABLES");
$tables = array();
while($row = mysql_fetch_row($getTables)){
$tables[] = $row[0];
}
$createTable = mysql_query("CREATE DATABASE `$newDbName` DEFAULT CHARACTER SET utf8 COLLATE utf8_general_ci;") or die(mysql_error());
foreach($tables as $cTable){
$db_check = @mysql_select_db ( $newDbName );
$create = $admin->query("CREATE TABLE $cTable LIKE ".$dbName.".".$cTable);
if(!$create) {
$error = true;
}
$insert = $admin->query("INSERT INTO $cTable SELECT * FROM ".$dbName.".".$cTable);
}
return !isset($error);
}
// usage
$clone = cloneDatabase('dbname','newdbname'); // first: toCopy, second: new database
实际上,我想在PHP中实现这一点,但这里没有一个答案是非常有用的,所以这里是我的-相当直接-使用MySQLi的解决方案:
// Database variables
$DB_HOST = 'localhost';
$DB_USER = 'root';
$DB_PASS = '1234';
$DB_SRC = 'existing_db';
$DB_DST = 'newly_created_db';
// MYSQL Connect
$mysqli = new mysqli( $DB_HOST, $DB_USER, $DB_PASS ) or die( $mysqli->error );
// Create destination database
$mysqli->query( "CREATE DATABASE $DB_DST" ) or die( $mysqli->error );
// Iterate through tables of source database
$tables = $mysqli->query( "SHOW TABLES FROM $DB_SRC" ) or die( $mysqli->error );
while( $table = $tables->fetch_array() ): $TABLE = $table[0];
// Copy table and contents in destination database
$mysqli->query( "CREATE TABLE $DB_DST.$TABLE LIKE $DB_SRC.$TABLE" ) or die( $mysqli->error );
$mysqli->query( "INSERT INTO $DB_DST.$TABLE SELECT * FROM $DB_SRC.$TABLE" ) or die( $mysqli->error );
endwhile;