mysql - 不使用 mysqldump 复制/复制数据库

Question

如果没有对服务器的本地访问，有没有办法在不使用的情况下将 MySQL 数据库（有内容和没有内容）复制/克隆到另一个数据库中mysqldump？

我目前正在使用 MySQL 4.0。

score 689 · Accepted Answer

我可以看到您说您不想使用mysqldump，但是我在寻找类似的解决方案时到达了此页面，其他人也可能会找到它。考虑到这一点，这里有一种从 Windows 服务器的命令行复制数据库的简单方法：

使用 MySQLAdmin 或您的首选方法创建目标数据库。在此示例中，db2是目标数据库，db1将复制源数据库。
在命令行上执行以下语句：

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

-p注意：和之间没有空格[password]

score 137 · Accepted Answer

您可以通过运行复制没有数据的表：

CREATE TABLE x LIKE y;

（参见MySQL CREATE TABLE文档）

您可以编写一个脚本，SHOW TABLES从一个数据库获取输出并将架构复制到另一个数据库。您应该能够引用架构+表名称，例如：

CREATE TABLE x LIKE other_db.y;

就数据而言，你也可以在 MySQL 中进行，但不一定很快。创建引用后，您可以运行以下命令来复制数据：

INSERT INTO x SELECT * FROM other_db.y;

如果你使用 MyISAM，你最好复制表文件；它会快得多。如果您将 INNODB 与per table table spaces一起使用，您应该也可以这样做。

如果你最终做了一个，INSERT INTO SELECT一定要暂时关闭索引ALTER TABLE x DISABLE KEYS！

EDIT Maatkit也有一些脚本可能有助于同步数据。它可能不会更快，但您可以在没有太多锁定的情况下在实时数据上运行他们的同步脚本。

score 60 · Accepted Answer

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，但它可以工作......而且它比 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

score 13 · Accepted Answer

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

score 4 · Accepted Answer

注意有一个 mysqldbcopy 命令作为添加 mysql 实用程序的一部分...... https://dev.mysql.com/doc/mysql-utilities/1.5/en/utils-task-clone-db.html

score 2 · Accepted Answer

所有先前的解决方案都有一点点，但是，它们只是不会复制所有内容。我创建了一个 PHP 函数（虽然有点长），它复制所有内容，包括表、外键、数据、视图、过程、函数、触发器和事件。这是代码：

/* 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]));
        }
    }
}

score 2 · Accepted Answer

实际上，我想在 PHP 中完全实现这一点，但这里的答案都不是很有帮助，所以这是我的 - 非常简单 - 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;

score 1 · Accepted Answer

我真的不知道您所说的“本地访问”是什么意思。但是对于该解决方案，您需要能够通过 ssh 访问服务器以复制存储数据库的文件。

我不能使用mysqldump，因为我的数据库很大（7Go，mysqldump失败）如果2个mysql数据库的版本差异太大可能无法使用，您可以使用mysql -V检查您的mysql版本。

1）将数据从远程服务器复制到本地计算机（vps是远程服务器的别名，可以替换为root@1.2.3.4）

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）导入本地计算机上复制的数据

/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

如果您有不同的版本，您可能需要运行

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

score 1 · Accepted Answer

在没有 mysqldump 的情况下克隆数据库表的最佳方法：

创建一个新的数据库。

使用查询创建克隆查询：

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';

运行那个输出！

但请注意，上面的脚本只是快速克隆表- 而不是视图、触发器和用户函数：您可以通过快速获取结构mysqldump --no-data --triggers -uroot -ppassword，然后使用仅克隆 insert 语句。

为什么这是实际问题？因为如果数据库超过 2Gb ，mysqldumps 的上传速度会很慢。而且您不能仅通过复制数据库文件（如快照备份）来克隆 InnoDB 表。

score 0 · Accepted Answer

显示 SQL 命令的 SQL，需要运行以将数据库从一个数据库复制到另一个数据库。对于每个表，都有一个创建表语句和一个插入语句。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';

score -2 · Accepted Answer

mysqldump 是不错的解决方案。复制数据库的最简单方法：

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

此外，您可以通过这种方式更改存储引擎：

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

mysql - 不使用 mysqldump 复制/复制数据库

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