当表很大时，最好从控制台进行操作

convert-to-innodb.sh

#!/usr/bin/env bash

# Usage: ./convert-to-innodb.sh 'db' 'user' 'password' | mysql 'db' -u  user -p password


set -eu

db="$1"
user="$2"
pass="$3"


sql="SET @DATABASE_NAME = '${db}';"

sql+="SELECT  CONCAT('ALTER TABLE \`', table_name, '\` ENGINE=InnoDB;') AS sql_statements
FROM    information_schema.tables AS tb
WHERE   table_schema = @DATABASE_NAME
AND     \`ENGINE\` = 'MyISAM'
AND     \`TABLE_TYPE\` = 'BASE TABLE'
ORDER BY table_name DESC;"

echo $sql | mysql -u${user} -p${pass} | tail -n +2 

您可以用您最喜欢的脚本语言编写一个脚本来完成它。该脚本将执行以下操作:

Issue显示满表。对于返回的每一行，检查第二列是否显示为“BASE TABLE”而不是“VIEW”。如果它不是'VIEW'，发出适当的ALTER TABLE命令。

<?php
    // connect your database here first 
    // 

    // Actual code starts here 

    $sql = "SELECT TABLE_NAME FROM INFORMATION_SCHEMA.TABLES
        WHERE TABLE_SCHEMA = 'your_database_name' 
        AND ENGINE = 'MyISAM'";

    $rs = mysql_query($sql);

    while($row = mysql_fetch_array($rs))
    {
        $tbl = $row[0];
        $sql = "ALTER TABLE `$tbl` ENGINE=INNODB";
        mysql_query($sql);
    }
?>

<?php

  // connect your database here first

  mysql_connect('host', 'user', 'pass');

  $databases = mysql_query('SHOW databases');

  while($db = mysql_fetch_array($databases)) {
    echo "database => {$db[0]}\n";
    mysql_select_db($db[0]);

    $tables = mysql_query('SHOW tables');

    while($tbl = mysql_fetch_array($tables)) {
      echo "table => {$tbl[0]}\n";
      mysql_query("ALTER TABLE {$tbl[0]} ENGINE=InnoDB");
    }
  }

你可以在mysql命令行工具中执行这个语句:

echo "SELECT concat('ALTER TABLE `',TABLE_NAME,'` ENGINE=InnoDB;')
FROM Information_schema.TABLES 
WHERE ENGINE != 'InnoDB' AND TABLE_TYPE='BASE TABLE' 
AND TABLE_SCHEMA='name-of-database'" | mysql > convert.sql

您可能需要指定用户名和密码:mysql -u username -p 结果是一个sql脚本，你可以管道回mysql:

mysql name-of-database < convert.sql

替换上述语句和命令行中的"name-of-database"。

使用这一行更改单个表的数据库引擎。

  ALTER TABLE table_name ENGINE = INNODB;