【发布时间】:2016-05-23 01:09:39
【问题描述】:
所以我有一个用于连接数据库的帮助类,但现在我希望能够使用同一个类在同一个代码块中连接到不同的数据库。
助手类:
<?php
require_once 'config.php'; // Database setting constants [DB_HOST, DB_NAME, DB_USERNAME, DB_PASSWORD]
class dbHelper {
private $db;
private $err;
function __construct() {
$dsn = 'mysql:host='.DB_HOST.';dbname='.DB_NAME.';charset=utf8';
try {
$this->db = new PDO($dsn, DB_USERNAME, DB_PASSWORD, array(PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION));
} catch (PDOException $e) {
$response["status"] = "error";
$response["message"] = 'Connection failed: ' . $e->getMessage();
$response["data"] = null;
exit;
}
}
function select($table, $where){
try{
$a = array();
$w = "";
foreach ($where as $key => $value) {
$w .= " and " .$key. " like :".$key;
$a[":".$key] = $value;
}
$stmt = $this->db->prepare("select * from ".$table." where 1=1 ". $w);
$stmt->execute($a);
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
if(count($rows)<=0){
$response["status"] = "warning";
$response["message"] = "No data found.";
}else{
$response["status"] = "success";
$response["message"] = "Data selected from database";
}
$response["data"] = $rows;
}catch(PDOException $e){
$response["status"] = "error";
$response["message"] = 'Select Failed: ' .$e->getMessage();
$response["data"] = null;
}
return $response;
}
}
所以你可以通过设置常量然后调用函数来调用上面的代码:
dbconfig.php
<?php
/**
* Database configuration
*/
define('DB_USERNAME', 'root');
define('DB_PASSWORD', '');
define('DB_HOST', 'localhost');
define('DB_NAME', 'database1');
?>
page.php
<?php
require_once 'dbHelper.php';
$db = new dbHelper();
$rows = $db->select("customers_php",array());
print_r(json_encode($rows,JSON_NUMERIC_CHECK));
?>
我想放弃对 config.php 文件的需求,并将 DB_HOST、DB_NAME 等移动到类内部的函数中,并将数据库名称与 $table 和 $where 信息一起传递。
if($db_name == 'database1')
{
//#----------open database connection --------------------> TESTING
$db_host = "localhost";
$db_user = "root";
$db_password = "";
$db_name = "database1";
}
if($db_name == 'database2')
{
//#----------open database connection --------------------> TESTING
$db_host = "localhost";
$db_user = "root";
$db_password = "";
$db_name = "database2";
}
newpage.php
<?php
require_once 'dbHelper.php';
$db = new dbHelper();
//function select($dbname, $table, $where)......................
$rows = $db->select("database1","customers_php",array());
print_r(json_encode($rows,JSON_NUMERIC_CHECK));
?>
那么,我应该/可以将“if($dbname)”部分放在哪里?
【问题讨论】:
-
哪些数据库,即 mysql - 其他?数据库是否在同一台服务器上?你想
query them against each other吗?我只是好奇,因为您已经接受了您认为有用的答案。 -
数据库在同一台服务器上,现在无论如何。如果一个变得相当大,我们可能会考虑将它移到它自己的服务器上。
-
不,此时不需要相互查询。
-
感谢您的澄清 - 它有帮助。