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复制/复制数据库而不使用mysqldump

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

我目前正在使用MySQL 4.0 .

9 回答

  • 2

    我可以看到你说你不想使用 mysqldump ,但我在寻找类似的解决方案时到达了这个页面,其他人也可能会找到它 . 考虑到这一点,这是从Windows服务器的命令行复制数据库的简单方法:

    • 使用MySQLAdmin或您首选的方法创建目标数据库 . 在此示例中, db2 是目标数据库,其中将复制源数据库 db1 .

    • 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

    注意: -p[password] 之间没有空格

  • 681

    您可以通过运行来复制没有数据的表:

    CREATE TABLE x LIKE y;
    

    (见MySQL CREATE TABLE文件)

    您可以编写一个脚本,从一个数据库获取 SHOW TABLES 的输出,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,如:

    CREATE TABLE x LIKE other_db.y;
    

    就数据而言,你也可以在MySQL中做到这一点,但它并不一定快 . 创建引用后,可以运行以下命令来复制数据:

    INSERT INTO x SELECT * FROM other_db.y;
    

    如果你're using MyISAM, you'最好复制表文件;它'll be much faster. You should be able to do the same if you'重新使用INNODB与per table table spaces .

    如果你最终做了 INSERT INTO SELECT ,请务必暂时turn off indexesALTER TABLE x DISABLE KEYS

    EDIT Maatkit也有一些可能有助于同步数据的脚本 . 它可能不会更快,但你可以在没有太多锁定的情况下在实时数据上运行它们的同步脚本 .

  • 58

    如果您使用的是Linux,则可以使用此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
    
  • 133

    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
    
  • 2

    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

  • 0

    我真的不知道你的意思"local access" . 但是对于该解决方案,您需要能够通过服务器ssh访问 copy the files where is database is stored .

    我不能使用mysqldump,因为我的数据库很大(7Go,mysqldump失败)如果2 mysql数据库的版本太不同它可能不起作用,你可以使用mysql -V检查你的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,可以用root@1.2.3.4替换)

    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,则可能需要运行

    /etc/init.d/mysql stop
    mysql_upgrade -u root -pPASSWORD --force #that step took almost 1hrs
    /etc/init.d/mysql start
    
  • 0

    所有先前的解决方案都达到了一点,然而,他们只是没有复制一切 . 我创建了一个PHP函数(尽管有点冗长),它复制了包括表,外键,数据,视图,过程,函数,触发器和事件在内的所有内容 . 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]));
            }
        }
    }
    
  • 11

    在没有mysqldump的情况下克隆数据库表的最佳方法:

    • 创建一个新数据库 .

    • 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';
    
    • 运行该输出!

    但请注意,上面的脚本只是 fast clone tables - 不是视图,触发器和用户函数:您可以通过 mysqldump --no-data --triggers -uroot -ppassword 快速获取结构,然后使用仅克隆插入语句 .

    为什么这是实际问题?因为 uploading of mysqldumps is ugly slow 如果DB超过2Gb . 而且您无法通过复制数据库文件(如快照备份)来克隆InnoDB表 .

  • 1

    创建SQL命令以复制行:

    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';
    

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