php - 如何在 PHP 中使用类连接两个 Mysql 数据库

Question

我正在使用面向对象的 PHP。为了连接同一主机中的两个数据库，我使用这样的代码

<?php
class ConOs{
    var $dbhost1;
    var $dbuser1;
    var $dbpass1;
    var $dbhost2;
    var $dbuser2;
    var $dbpass2;

    function __construct() {

                            }
                            function connect(){
        $this->dbhost1="localhost";
        $this->dbuser1="user1";
        $this->dbpass1="db1";
        $this->dbhost2="localhost";
        $this->dbuser2="user2";
        $this->dbpass2="db2";

$conn1 = mysql_connect($this->dbhost1, $this->dbuser1, $this->dbpass1);
if(! $conn1 ){die('Could not connect: ' . mysql_error());}
mysql_select_db("db1",$conn1);


$conn2 = mysql_connect($this->dbhost2, $this->dbuser2, $this->dbpass2, True);
if(! $conn2 ){die('Could not connect: ' . mysql_error());}
mysql_select_db("db2", $conn2);


mysql_error();
                            }
}
?>

但是 $conn1 总是拒绝。只有 $conn2 有效。如果我把 $conn2 首先，那么只有 $conn1 有效。如何连接数据库 db1 和 db2？

Answer 1

亲爱的停止使用 mysql_* 函数:使用 PDO :

class sujet{
private $id;

public function connect(){          
    try {
            $bdd = new PDO('mysql:host=localhost;dbname=dev_bdd', 'root', '');
    }catch(Exception $e){
            die('Failed'. $e->getMessage());
    }
    return $bdd;            
}       
public function connect2(){         
    try {
            $bdd = new PDO('mysql:host=localhost;dbname=teamd2', 'root', '');
    }catch(Exception $e){
            die('Failed'. $e->getMessage());
    }
    return $bdd;            
}
public function __construct(){} 
public function recherche($search){
    $resultats = array();
    $req = $this->connect2()->prepare('SELECT * FROM sujets WHERE libelle like :search ');
    $req->execute(array(
        'search' => '%'.$search.'%',
         ));
    while($resultat = $req->fetch()){
        array_push($resultats,$resultat);
    }
    print_r($resultats);
    return $resultats;      
}
    private function queryMatiere(int $id){
    $requete = $this->connect()->query("SELECT libelle FROM matieres where id=$id");
    $data = $requete->fetch();
    return $data['libelle'];
}
}