Copy/duplicate database without using mysqldump


Without local access to the server, is there any way to duplicate/clone a MySQL db (with content and without content) into another without using mysqldump?

I am currently using MySQL 4.0.

This question is tagged with mysql

~ Asked on 2008-08-25 08:47:27

11 Answers


I can see you said you didn't want to use mysqldump, but I reached this page while looking for a similar solution and others might find it as well. With that in mind, here is a simple way to duplicate a database from the command line of a windows server:

  1. Create the target database using MySQLAdmin or your preferred method. In this example, db2 is the target database, where the source database db1 will be copied.
  2. Execute the following statement on a command line:

mysqldump -h [server] -u [user] -p[password] db1 | mysql -h [server] -u [user] -p[password] db2

Note: There is NO space between -p and [password]

~ Answered on 2011-08-18 17:00:30


You can duplicate a table without data by running:



You could write a script that takes the output from SHOW TABLES from one database and copies the schema to another. You should be able to reference schema+table names like:

CREATE TABLE x LIKE other_db.y;

As far as the data goes, you can also do it in MySQL, but it's not necessarily fast. After you've created the references, you can run the following to copy the data:


If you're using MyISAM, you're better off to copy the table files; it'll be much faster. You should be able to do the same if you're using INNODB with per table table spaces.

If you do end up doing an INSERT INTO SELECT, be sure to temporarily turn off indexes with ALTER TABLE x DISABLE KEYS!

EDIT Maatkit also has some scripts that may be helpful for syncing data. It may not be faster, but you could probably run their syncing scripts on live data without much locking.

~ Answered on 2008-08-25 14:19:24


If you are using Linux, you can use this bash script: (it perhaps needs some additional code cleaning but it works ... and it's much faster then mysqldump|mysql)



echo "Copying database ... (may take a while ...)"
DBCONN="-h ${DBSERVER} -u ${DBUSER} --password=${DBPASSWORD}"
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} ; ${createTable}"
        fInsertData="${fInsertData} ; ${insertData}"
echo "$fCreateTable ; $fInsertData" | mysql $DBCONN $DBNAME

~ Answered on 2010-01-03 14:54:11



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

~ Answered on 2011-08-31 11:33:21


Note there is a mysqldbcopy command as part of the add on mysql utilities....

~ Answered on 2017-05-25 13:15:02


Actually i wanted to achieve exactly that in PHP but none of the answers here were very helpful so here's my – pretty straightforward – solution using MySQLi:

// 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 );


~ Answered on 2019-09-13 06:20:22


All of the prior solutions get at the point a little, however, they just don't copy everything over. I created a PHP function (albeit somewhat lengthy) that copies everything including tables, foreign keys, data, views, procedures, functions, triggers, and events. Here is the code:

/* This function takes the database connection, an existing database, and the new database and duplicates everything in the new database. */
function copyDatabase($c, $oldDB, $newDB) {

    // creates the schema if it does not exist
    $schema = "CREATE SCHEMA IF NOT EXISTS {$newDB};";
    mysqli_query($c, $schema);

    // selects the new schema
    mysqli_select_db($c, $newDB);

    // gets all tables in the old schema
    $tables = "SELECT table_name
               FROM information_schema.tables
               WHERE table_schema = '{$oldDB}'
               AND table_type = 'BASE TABLE'";
    $results = mysqli_query($c, $tables);

    // checks if any tables were returned and recreates them in the new schema, adds the foreign keys, and inserts the associated data
    if (mysqli_num_rows($results) > 0) {

        // recreates all tables first
        while ($row = mysqli_fetch_array($results)) {
            $table = "CREATE TABLE {$newDB}.{$row[0]} LIKE {$oldDB}.{$row[0]}";
            mysqli_query($c, $table);

        // resets the results to loop through again
        mysqli_data_seek($results, 0);

        // loops through each table to add foreign key and insert data
        while ($row = mysqli_fetch_array($results)) {

            // inserts the data into each table
            $data = "INSERT IGNORE INTO {$newDB}.{$row[0]} SELECT * FROM {$oldDB}.{$row[0]}";
            mysqli_query($c, $data);

            // gets all foreign keys for a particular table in the old schema
            $fks = "SELECT constraint_name, column_name, table_name, referenced_table_name, referenced_column_name
                    FROM information_schema.key_column_usage
                    WHERE referenced_table_name IS NOT NULL
                    AND table_schema = '{$oldDB}'
                    AND table_name = '{$row[0]}'";
            $fkResults = mysqli_query($c, $fks);

            // checks if any foreign keys were returned and recreates them in the new schema
            // Note: ON UPDATE and ON DELETE are not pulled from the original so you would have to change this to your liking
            if (mysqli_num_rows($fkResults) > 0) {
                while ($fkRow = mysqli_fetch_array($fkResults)) {
                    $fkQuery = "ALTER TABLE {$newDB}.{$row[0]}                              
                                ADD CONSTRAINT {$fkRow[0]}
                                FOREIGN KEY ({$fkRow[1]}) REFERENCES {$newDB}.{$fkRow[3]}({$fkRow[1]})
                                ON UPDATE CASCADE
                                ON DELETE CASCADE;";
                    mysqli_query($c, $fkQuery);

    // gets all views in the old schema
    $views = "SHOW FULL TABLES IN {$oldDB} WHERE table_type LIKE 'VIEW'";                
    $results = mysqli_query($c, $views);

    // checks if any views were returned and recreates them in the new schema
    if (mysqli_num_rows($results) > 0) {
        while ($row = mysqli_fetch_array($results)) {
            $view = "SHOW CREATE VIEW {$oldDB}.{$row[0]}";
            $viewResults = mysqli_query($c, $view);
            $viewRow = mysqli_fetch_array($viewResults);
            mysqli_query($c, preg_replace("/CREATE(.*?)VIEW/", "CREATE VIEW", str_replace($oldDB, $newDB, $viewRow[1])));

    // gets all triggers in the old schema
    $triggers = "SELECT trigger_name, action_timing, event_manipulation, event_object_table, created
                 FROM information_schema.triggers
                 WHERE trigger_schema = '{$oldDB}'";                 
    $results = mysqli_query($c, $triggers);

    // checks if any triggers were returned and recreates them in the new schema
    if (mysqli_num_rows($results) > 0) {
        while ($row = mysqli_fetch_array($results)) {
            $trigger = "SHOW CREATE TRIGGER {$oldDB}.{$row[0]}";
            $triggerResults = mysqli_query($c, $trigger);
            $triggerRow = mysqli_fetch_array($triggerResults);
            mysqli_query($c, str_replace($oldDB, $newDB, $triggerRow[2]));

    // gets all procedures in the old schema
    $procedures = "SHOW PROCEDURE STATUS WHERE db = '{$oldDB}'";
    $results = mysqli_query($c, $procedures);

    // checks if any procedures were returned and recreates them in the new schema
    if (mysqli_num_rows($results) > 0) {
        while ($row = mysqli_fetch_array($results)) {
            $procedure = "SHOW CREATE PROCEDURE {$oldDB}.{$row[1]}";
            $procedureResults = mysqli_query($c, $procedure);
            $procedureRow = mysqli_fetch_array($procedureResults);
            mysqli_query($c, str_replace($oldDB, $newDB, $procedureRow[2]));

    // gets all functions in the old schema
    $functions = "SHOW FUNCTION STATUS WHERE db = '{$oldDB}'";
    $results = mysqli_query($c, $functions);

    // checks if any functions were returned and recreates them in the new schema
    if (mysqli_num_rows($results) > 0) {
        while ($row = mysqli_fetch_array($results)) {
            $function = "SHOW CREATE FUNCTION {$oldDB}.{$row[1]}";
            $functionResults = mysqli_query($c, $function);
            $functionRow = mysqli_fetch_array($functionResults);
            mysqli_query($c, str_replace($oldDB, $newDB, $functionRow[2]));

    // selects the old schema (a must for copying events)
    mysqli_select_db($c, $oldDB);

    // gets all events in the old schema
    $query = "SHOW EVENTS
              WHERE db = '{$oldDB}';";
    $results = mysqli_query($c, $query);

    // selects the new schema again
    mysqli_select_db($c, $newDB);

    // checks if any events were returned and recreates them in the new schema
    if (mysqli_num_rows($results) > 0) {
        while ($row = mysqli_fetch_array($results)) {
            $event = "SHOW CREATE EVENT {$oldDB}.{$row[1]}";
            $eventResults = mysqli_query($c, $event);
            $eventRow = mysqli_fetch_array($eventResults);
            mysqli_query($c, str_replace($oldDB, $newDB, $eventRow[3]));

~ Answered on 2018-01-08 05:08:14


I don't really know what you mean by "local access". But for that solution you need to be able to access over ssh the server to copy the files where is database is stored.

I cannot use mysqldump, because my database is big (7Go, mysqldump fail) If the version of the 2 mysql database is too different it might not work, you can check your mysql version using mysql -V.

1) Copy the data from your remote server to your local computer (vps is the alias to your remote server, can be replaced by [email protected])

ssh vps:/etc/init.d/mysql stop
scp -rC vps:/var/lib/mysql/ /tmp/var_lib_mysql
ssh vps:/etc/init.d/apache2 start

2) Import the data copied on your local computer

/etc/init.d/mysql stop
sudo chown -R mysql:mysql /tmp/var_lib_mysql
sudo nano /etc/mysql/my.cnf
-> [mysqld]
-> datadir=/tmp/var_lib_mysql
/etc/init.d/mysql start

If you have a different version, you may need to run

/etc/init.d/mysql stop
mysql_upgrade -u root -pPASSWORD --force #that step took almost 1hrs
/etc/init.d/mysql start

~ Answered on 2017-05-30 08:17:10


The best way to clone database tables without mysqldump:

  1. Create a new database.
  2. Create clone-queries with query:

    SET @NewSchema = 'your_new_db';
    SET @OldSchema = 'your_exists_db';
    SELECT CONCAT('CREATE TABLE ',@NewSchema,'.',table_name, ' LIKE ', TABLE_SCHEMA ,'.',table_name,';INSERT INTO ',@NewSchema,'.',table_name,' SELECT * FROM ', TABLE_SCHEMA ,'.',table_name,';') 
    FROM information_schema.TABLES where TABLE_SCHEMA = @OldSchema AND TABLE_TYPE != 'VIEW';
  3. Run that output!

But note, script above just fast clone tables - not views, triggers and user-functions: you can fast get structure by mysqldump --no-data --triggers -uroot -ppassword , and then use to clone only insert statement .

Why it is actual question? Because uploading of mysqldumps is ugly slow if DB is over 2Gb. And you can't clone InnoDB tables just by copying DB files (like snapshot backuping).

~ Answered on 2017-12-10 16:06:01


an SQL that shows SQL commands, need to run to duplicate a database from one database to another. for each table there is create a table statement and an insert statement. it assumes both databases are on the same server:

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,"`;") 


as sqlstatement
 FROM information_schema.tables where [email protected] and TABLE_TYPE='BASE TABLE';

~ Answered on 2019-03-20 09:01:26


Mysqldump isn't bad solution. Simplest way to duplicate database:

mysqldump -uusername -ppass dbname1 | mysql -uusername -ppass dbname2

Also, you can change storage engine by this way:

mysqldump -uusername -ppass dbname1 | sed 's/InnoDB/RocksDB/' | mysql -uusername -ppass dbname2

~ Answered on 2020-01-15 11:33:25

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