php - 创建具有多个 db 信息的 db conn 文件

Question

我有 4 个不同的数据库，有 4 个不同的用户名和密码。我想把所有的连接信息放到一个文件中：db_conn.php。

db_conn.php

<?php

$dbname1 = "courses_classes";
     $web_name ='wart.wmysql.com';
     $web_user = "admin";
     $web_pass="administrator";

$dbname2 ="courses_supplies";
     $web_name ='wart.wmysql.com';
     $web_user = "maint";
     $web_pass="maintenance";
     
$dbname3 = "courses_instructor":
     $web_name ='wart.wmysql.com';
     $web_user = "teacher";
     $web_pass="instructor";

$dbname4 = "webmaster";
     $web_name ='wart.wmysql.com';
     $web_user = "admin";
     $web_pass="webAdmin";
     
?>

在不同的页面上，如果可能的话，这是我想使用的代码：

//============================================================ database connection
require('../cgi-bin/db_conn.php');

$dbname = "dbname1";
$table_name = "members";

// Create connection
$conn= new mysqli($web_name,$web_user ,$web_pass,$dbname);

// Check connection
if ($conn-> connect_errno) {
     echo "Failed to connect to MySQL: " . $conn-> connect_error;
     exit();
} 
//else { echo "connected"; }
// ============================================================= end db connection

标签： phpmysqlidatabase-connection

Answer 1

不让 conn 为空，做了一点小改动。这是现在的样子

连接页面

$db = [
        'admin' => [
             'db_conn' => 'schoolnet.qwertymysql.com',
             'db_user'  => 'user',
             'db_pass'  => 'pass',
             'db_name' => 'school_comm',
             ],
];

网页

    $selectedConn = 'Content';
    $conn= new mysqli($db[$selectedConn]['db_conn'],
                      $db[$selectedConn]['db_user'] ,
                      $db[$selectedConn]['db_pass'],
                      $db[$selectedConn]['db_name']
                      );