如何在PHP MYSQL ?>中将数据库表和每条记录从一台数据库服务器复制到另一台数据库服务器

Question

嗨，我写了一个代码，可以将数据库表从一个服务器复制到另一个服务器，但每个表的记录都没有复制。如何编写一个函数，可以将表和每个记录从一个数据库服务器复制到另一个数据库服务器？

下面是我的示例代码：

<?php
$dbNewDB = 'newdb';
$dbNewUser = 'newroot';
$dbNewUserPswd = 'newpass';

$dbConnect = mysql_connect('localhost', 'root', 'mypassword') or die('Couldn\'t connect to MySql:'.mysql_error());
$dbNewConnect = mysql_connect('localhost', $dbNewUser, $dbNewUserPswd) or die('Couldn\'t connect to MySql:'.mysql_error());

$sqlStatement = "SHOW TABLES FROM olddb";
$result = mysql_query($sqlStatement,$dbConnect) or die('Unable to get tables: '.mysql_error());
while($row = mysql_fetch_row($result)) 
    {
        //Drop table if exist
        $sqlStatement = "DROP TABLE IF EXISTS " . $dbNewDB . "." . $row[0];
        mysql_query($sqlStatement,$dbNewConnect) or die("Failed to delete: " . mysql_error());
        //Create new table
        $sqlStatement = "CREATE TABLE " . $dbNewDB . "." . $row[0] . " LIKE olddb." . $row[0];
        echo "$sqlStatement [" . __METHOD__ . "]"; 
        mysql_query($sqlStatement,$dbNewConnect)or die("Failed to create: ". mysql_error());
        //Insert data
        $sqlStatement = "INSERT INTO " . $dbNewDB . "." . $row[0] . " SELECT * FROM " . $dbNewDB . "." . $row[0];
        echo "$sqlStatement [" . __METHOD__ . "]"; 
        mysql_query($sqlStatement,$dbNewConnect)or die("Table copy failed: ".mysql_error());
        echo "$row[0] copy done. [" . __METHOD__ . "]"; 
    }

mysql_free_result($result);
mysql_close($dbConnect);
mysql_close($dbNewConnect);

?> 

我的代码已经是功能性的，我想要修复的是复制每个表的记录。有什么想法吗?或者帮帮忙？

谢谢!

Answer 1

您可以使用SQL格式转储整个数据库，如下所示：

mysqldump --user=root --password=whatever --databases dbtest --opt --quote-names --complete-insert > testbkup.sql

然后，您可以像这样将其导入回去：

mysql -u root -p whatever dbtest < testbkup.sql

(注意: user = root，password =任何值，dbtest是您的数据库。)

只是说说而已。

Answer 2

我发现这些脚本很有效，你可以试试这些：

<?php
$connect2 = mysql_connect("localhost", "root", "");
$database1 = "test1"; // destination database
mysql_select_db($database1, $connect2);
set_time_limit(0);

$database = 'dev_loribonn'; //original database
$connect = mysql_connect("localhost", "root", "");

mysql_select_db($database, $connect);

$tables = mysql_query("SHOW TABLES FROM $database");

while ($line = mysql_fetch_row($tables)) {
    $tab = $line[0];
    mysql_query("DROP TABLE IF EXISTS $database1.$tab");
    mysql_query("CREATE TABLE $database1.$tab LIKE $database.$tab") or die(mysql_error());
    mysql_query("INSERT INTO $database1.$tab SELECT * FROM $database.$tab");
    echo "Table: <b>" . $line[0] . " </b>Done<br>";
}
?>

Answer 3

您的insert语句看起来不太对劲。尝试使用来自dbNewDB的值插入到$dbNewDB中。您必须转向旧的数据库。下面我正在为插入构建两个stings。$string1 = '(col1name，col2name...，)‘$string2 = '(val1-1，val1-2，...)，(val2-1，val2-2，...)，...’对于"INSERT for $string1 VALUES $string2“

    //Insert data 

    $sql2 = "SELECT * FROM " . $row[0];
    $r = mysql_query($sql, $bConnect);
    $string1 = '('; $arr = array();
    while ($irow = mysql_fetch_assoc($r)) {$arr[] = $irow;}
    foreach($irow as $k=>$v)
    {
         $string1 .= "$k,";
    }
    $string1 = substr($string1, 0, -1) //lose last comma
    $string1 .= ')';
    $string2 = array_reduce($f, $arr); 
    $string2 = substr($string2, 0, -1) //lose last comma


    $sqlStatement = "INSERT INTO " . $dbNewDB . "." . $row[0] . " $string1 VALUES $string2";
    echo "$sqlStatement [" . __METHOD__ . "]";  
    mysql_query($sqlStatement,$dbNewConnect)or die("Table copy failed: ".mysql_error()); 
    echo "$row[0] copy done. [" . __METHOD__ . "]";  

在其他地方声明$f

$f = function($b, $x) {$a = ' ('; foreach($x as $v) {$a .= "'$v',";} $a = substr($a, 0, -1); $a .= ')'; return "$b $a,";}