我正在尝试使用运行在MySQL服务器上的PHP脚本,在一个新的数据库中复制一个数据库。目前为止,我所拥有的代码如下:
$dbh->exec("CREATE DATABASE IF NOT EXISTS $new_news CHARACTER SET UTF8;");
$results = $dbh->query("SHOW TABLES FROM $old_news");
$table_list = $results->fetchAll(PDO::FETCH_NUM);
foreach($table_list as $table_row){
foreach($table_row as $table){
$results = $dbh->query("SELECT table_type FROM information_schema.tables where table_schema = '$old_news' and table_name = '$table'");
$table_type = $results->fetch(PDO::FETCH_ASSOC);
$table_type = $table_type['table_type'];
if($table_type == 'BASE TABLE'){
echo "Creating table $table and populating...\n";
$dbh->exec("CREATE TABLE $new_news.$table LIKE $old_news.$table");
$dbh->exec("INSERT INTO $new_news.$table SELECT * FROM $old_news.$table");
}else if($table_type == 'VIEW'){
//echo "Creating view $table...\n";
//$dbh->exec("CREATE VIEW $new_news.$table LIKE $old_news.$table");
echo "$table is a view, which cannot be copied atm\n";
}else{
echo "Skipping $table_type $table, unsupported type\n";
}
}
}
目前这个程序会查找$old_news中的所有表,通过information_schema找到表类型,并在$new_news中创建一个相同类型的表。对于表格,它会创建相同的表结构,然后使用“INSERT INTO SELECT”来填充它们。
如何在不使用mysqldump备份整个数据库的情况下复制视图?