如果主数据库关闭，如何连接到辅助数据库（PHP） [英] how to conect to secondary database if primary database is down (PHP)

问题描述

我希望有人能帮助我，我已经为此努力了3天。

i hope someone can help me, ive been struggling on this for 3 days now.

这是我的情况，我正在用php创建一个网站，并且我有2台计算机作为服务器与wampserver ...

here is my situation, i am making a website with php, and i have 2 computers as servers with wampserver...

主服务器为192.168.0.10

main server is 192.168.0.10

辅助服务器为192.168.0.12

secondary server is 192.168.0.12

我尝试远程连接是否正常的机器

and a virtual machine where im trying out if the remote conection works

我的网站托管在我的主服务器上，因此conexion查询是...

my website is hosted on my main server so the conexion query is...

$conexion = mysqli_connect("localhost","root","","dbdaq");

工作正常，我什至拥有复制服务器上的复制主机。

it works fine, i even have master to master replication on the servers.

但是我需要做的是，当主服务器关闭时，它需要尝试连接到seondary数据库上的数据库，所以我试图使用代码...

but what i need to do is that when main server is down it needs to try to conect to the database on the seondary database, so im trying to use the code...

$conexion = mysqli_connect("localhost","root","","dbdaq");


if (!$conexion){
    $conexion = mysqli_connect("192.168.0.12","root","password","dbdaq");
}

但是当我手动打开主服务器上的mysql服务时，它实际上并没有尝试来使用其他服务器数据库...

but when i manually turn of the mysql services on main server it doesnt actually try to use the other servers database...

它一直给我说错误

警告：mysqli_connect （）：（HY000 / 2002）：由于目标计算机主动拒绝连接，因此无法建立连接。在C：\wamp64\www\PaginaV2\Pagina\PHP和JS\Procesoalumno.php在第2行上

Warning: mysqli_connect(): (HY000/2002): No connection could be made because the target machine actively refused it. in C:\wamp64\www\PaginaV2\Pagina\PHP y JS\Procesoalumno.php on line 2

推荐答案

尝试一下：使用php的 @ 运算符（错误控制操作符），以抑制第一次连接失败时出现的错误消息。

Try this: use php's @ operator (the error control operator) to suppress the error message from the first connection failure. Try this.

$conexion = @mysqli_connect("localhost","root","","dbdaq");

if ( !$conexion ) {
    $conexion = mysqli_connect("192.168.0.12","root","password","dbdaq");
}

故障转移连接的代码似乎正常工作，但仍然可以令人讨厌的错误消息。 @ 运算符禁止显示该消息。

It seems likely your code for the failover connection works correctly, but you still get an obnoxious error message. The @ operator suppresses the message.

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