在php脚本中创建数据库表的克隆
问题描述:
我必须创建数据库表的克隆。我与wordpress数据库合作,我想创建一个WP选项表的克隆。我如何创建数据库中的特定表的克隆。我发现一些代码在php脚本中创建数据库表的克隆
set_time_limit(0);
$username = 'XXXXXX';
$password = 'YYYYYY';
$hostname = 'ZZZZZZ';
$database = 'AAAAAA';
try {
$pdo = new PDO("mysql:host={$hostname};dbname={$database}", $username, $password);
}
catch(PDOException $e) {
die("Could not connect to the database\n");
}
echo '<pre>';
$stmt1 = $pdo->query('SHOW TABLES', PDO::FETCH_NUM);
foreach($stmt1->fetchAll() as $row) {
$stmt2 = $pdo->query("SHOW CREATE TABLE `$row[0]`", PDO::FETCH_ASSOC);
$table = $stmt2->fetch();
echo "{$table['Create Table']};\n\n";
}
echo '</pre>';
,但我不能明白,我怎么能我给哪条路径的文件夹中得到数据库。
答
我之前也有同样的问题,并从几小时的谷歌搜索结束以下代码。
希望这会有所帮助。
对于使用mysql_**
不推荐使用,我们很抱歉。
<?php
$DBIUser = 'someuser';
$DBIPass = 'thepassword';
$NewUser = 'someloser';
$NewPass = 'thepassword';
$oldServer = 'my crappy old mysql server domain';
$newServer = 'localhost';
if ($argv[0] > " ")
{
$dbname = $argv[1];
echo "Starting copy of the $argv[1] database.\n";
$dbpre = mysql_connect($oldServer, $DBIUser, $DBIPass);
mysql_select_db($dbname, $dbpre);
$sql = "SHOW TABLES FROM $dbname";
echo $sql."\n";
$result = mysql_query($sql);
if (!$result)
{
echo "DB Error, could not list tables\n";
echo 'MySQL Error: ' . mysql_error();
exit;
}
$dbtbl = mysql_connect($oldServer, $DBIUser, $DBIPass);
mysql_select_db($dbname, $dbpre);
$dbnew = mysql_connect($newServer, $NewUser, $NewPass);
mysql_select_db("mysql", $dbnew);
$res2 = mysql_query("CREATE DATABASE IF NOT EXISTS ".$dbname,$dbnew);
if (!$res2)
{
echo "DB Error, could not create database\n";
echo 'MySQL Error: ' . mysql_error();
exit;
}
mysql_select_db($dbname, $dbnew);
if($result === FALSE)
{
die(mysql_error());
}
$f = fopen($dbname.'.log', 'w');
fwrite($f, "Copy all tables in database $dbname on server $oldServer to new database on server $newServer.\n\n");
while ($row = mysql_fetch_row($result))
{
echo "Table: {$row[0]}\n";
fwrite($f, "Table ".$row[0]."\n");
$tableinfo = mysql_fetch_array(mysql_query("SHOW CREATE TABLE $row[0] ",$dbtbl));
$createsyntax = "CREATE TABLE IF NOT EXISTS ";
$createsyntax .= substr($tableinfo[1], 13);
mysql_query(" $createsyntax ",$dbnew);
$res = mysql_query("SELECT * FROM $row[0] ",$dbpre); // select all rows
$oldcnt = mysql_num_rows($res);
echo "Count: ".$oldcnt." - ";
$errors = 0;
while ($roz = mysql_fetch_array($res, MYSQL_ASSOC))
{
$query = "INSERT INTO $dbname.$row[0] (".implode(", ",array_keys($roz)).") VALUES (";
$cnt = 0;
foreach (array_values($roz) as $value)
{
if ($cnt == 0)
{
$cnt++;
} else
{
$query .= ",";
}
$query .= "'";
$query .= mysql_real_escape_string($value);
$query .= "'";
}
$query .= ")";
$look = mysql_query($query,$dbnew);
if ($look === false)
{
// write insert to log on error
$errors = $errors + 1;
fwrite($f, mysql_error()." - ".$query."\n");
}
}
$sql = "select count(*) as cnt from $dbname.$row[0] ";
$res = mysql_query($sql, $dbnew);
$roz = mysql_fetch_array($res);
echo $roz['cnt']." - Errors: ".$errors."\n";
fwrite($f, "Old Record Count: ".$oldcnt." - New Record Count: ".$roz['cnt']." - Errors: ".$errors."\n");
fwrite($f,"End table copy for table $row[0].\n\n");
}
fclose($f);
}
else
{
var_dump($argv);
}
?>
-1不推荐使用,因为这是非常糟糕的做法。 – Ben 2013-09-17 07:23:00
@Steve我知道,但你是否读过我写在第三行回答中的问题..... – 2013-09-17 07:40:23