php - 将来自 2 个数据库(A 和 B)的数据插入到数据库 B 的 1 个表中

Question

我正在尝试将来自 2 个不同数据库(A 和 B)中的 2 个不同表的数据插入到数据库 3 中的第 3 个表中。

这是我的代码:

// display all errors generated by script
error_reporting(E_ALL);
ini_set('display_errors','On');
include_once 'db_inc.php';  
try{
    $db = new PDO(DB_INFO, DB_USER, DB_PASS);
    //get the new customer array

    $query = $db->prepare("SELECT * FROM moodle");
    $query->execute(array());   
      $customer_arr=array();
      while($res= $query->fetch(PDO::FETCH_ASSOC)){
      $customer_arr[]=$res;
    }
}
catch (PDOException $e) {
    echo $err5="ERROR: ".$e->getMessage()."\n ";
}
try{
    $db = new PDO(DB_INFO4, DB_USER4, DB_PASS4);

    $nowtime = time();

    $sql4 = "INSERT INTO mdl_groups_members( groupid, userid , timeadded ) 
            SELECT l.mgroup AS moodle, r.id AS mdl_user
            FROM moodle l
            JOIN mdl_user r ON l.orders_id = r.id
            WHERE l.mgroup >0 ON DUPLICATE 
            KEY UPDATE groupid = VALUES (
            groupid
            )";
    $stmt4 = $db->prepare($sql4);
    $stmt4->execute(array($groupid,$userid,$nowtime));
}
catch (PDOException $e) {
    echo $err5="ERROR: ".$e->getMessage()."\n ";
}

我收到以下错误。

Notice: Undefined variable: groupid in ... on line 40

Notice: Undefined variable: userid in ... on line 40

任何帮助将不胜感激。谢谢。

Answer 1

你的最后一行(执行)崩溃了是的:
数组($groupid,$userid,$nowtime)您没有定义这 3 个变量...