如何将MySQL远程数据库表插入本地数据库表？

Question

远程服务器数据库(IP: 192.168.1.19)

数据库名称：RemoteDB 表名-相对较高的标准-比较：RemoteTable

本地服务器数据库(IP: 192.168.1.35)

数据库名称：LocalDB 表名-相对较高的标准-比较：LocalTable

MySQL查询

INSERT INTO LocalDB.LocalTable SELECT * FROM RemoteDB.RemoteTable

但是，由于不同的数据库连接，这个查询没有执行。

如何从本地服务器(IP: 192.168.1.35)执行此查询？或者任何linux shell命令？

Answer 1

mysqldump -uRemoteusername -pRemotepassword -h192.168.1.19 RemoteDB RemoteTable  --set-gtid-purged=OFF  | mysql -h192.168.1.35 -uLocalname -pLocalpassword   LocalDB

Answer 2

我将CronJob的批准答案转换为CronJob脚本，如下所示-

配置:

$remoteDbUser   = '***';
$remoteDbPass   = '***';
$remoteDbHost   = '192.168.1.19';
$remoteDb       = 'RemoteDB';
#$remoteDbTable = '';
$localDbHost    = '192.168.1.35';
$localDbUser    = '***';
$localDbPass    = '***';
$localDb        = 'LocalDB';

同步功能:

/*Passing `$remoteDbTable` name as function argument, Because I have many Remote DB table to synchronize with my Local DB table*/
function RemoteDbSynchronze($remoteDbTable='RemoteTable'){
  global $remoteDbUser,
    $remoteDbPass,
    $remoteDbHost,
    $remoteDb,
    $localDbHost,
    $localDbUser,
    $localDbPass,
    $localDb;
  $sql = sprintf(
    "mysqldump -u%s -p%s -h%s %s %s  --set-gtid-purged=OFF  | mysql -h%s -u%s -p%s %s",
    $remoteDbUser,
    $remoteDbPass,
    $remoteDbHost,
    $remoteDb,
    $remoteDbTable,
    $localDbHost,
    $localDbUser,
    $localDbPass,
    $localDb
  );
  shell_exec($sql);
}

从Cron脚本调用函数：

RemoteDbSynchronze('RemoteTable');
RemoteDbSynchronze('AnotherRemoteTable1');
RemoteDbSynchronze('AnotherRemoteTable2');
..........................................
RemoteDbSynchronze('AnotherRemoteTableN');

关于斯普林特、exec和克隆的更多详细信息

Answer 3

您的问题是使用php从不同的数据库中插入数据，不是吗？

如果你想同步它们，你可以读这个如何使两个MySQL数据库保持同步？

使用php从多个db插入db很简单

$conn1= mysqli_connect($host1,$user1,$pass1);
mysqli_select_db($db1,$conn1);
$conn2= mysqli_connect($host2,$user2,$pass2);
mysqli_select_db($db2,$conn2);

$query=mysql_query("SELECT id,name from tbl_user", $conn1)
while ($row = mysqli_fetch_row($query))
 {
    mysqli_query("insert into tbl_user (id,name) values ('".$row[0]."', '".$row[1]."',$conn2);
 }