問題描述

我可以創建一個 PDO 的實例并成功注入它.我直接定義了 PDO::class 并用 __construct(PDO $pdo) 將它注入到構造函數中.我需要像 PDO1::class 和 PDO2::class 之類的東西來注入它，如下所示:__construct(PDO1 $pdo1, PDO2 $pdo2) 但這顯然行不通.只有一個 PDO 類，我需要做的是它的 2 個具有不同數據庫憑據的實例.
最好的方法是什么?

I could make an instance of PDO and inject it successfully. I defined the PDO::class directly and injected it in the constructor with __construct(PDO $pdo). I would need something like PDO1::class and PDO2::class to inject it like follows: __construct(PDO1 $pdo1, PDO2 $pdo2) but that obviously doesn't work. There is only one PDO class and what I need to do is 2 instances of it with different database credentials.
What is the best way to do it?

我像這樣通過 PDO 設置了一個數據庫定義并且它可以工作:

I set up one definition of a database via PDO like this and it works:

文件:dependencies.php

use DIContainerBuilder;
use PsrContainerContainerInterface;

return function (ContainerBuilder $containerBuilder) {
    $containerBuilder->addDefinitions([
        PDO::class => function (ContainerInterface $c) {
            $dbSettings = $c->get('settings')['db1'];
            $dsn = 'mysql:host=' . $dbSettings['host'] . ';dbname=' . $dbSettings['dbname'];
            $options = [
                PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
                PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC,
                PDO::ATTR_EMULATE_PREPARES => false,
            ];
            return new PDO($dsn, $dbSettings['user'], $dbSettings['pass'], $options);
        },
    ]);
};

文件:index.php

...
// Set up dependencies
$dependencies = require __DIR__ . '/../app/dependencies.php';
$dependencies($containerBuilder);
// Build PHP-DI Container instance
$container = $containerBuilder->build();
// Set container to create App with on AppFactory
AppFactory::setContainer($container);
// Instantiate the app
$app = AppFactory::create();
...

文件 SomeRepository.php

use PDO;

class SomeRepository{

    protected $pdo;

    public function __construct(PDO $pdo) {
        $this->pdo = $pdo;
    }
}

我在這篇文章中看到過類似的內容:

I've seen something like this in this article:

return function (ContainerBuilder $containerBuilder) {
    $containerBuilder->addDefinitions([
        'db1' => function (ContainerInterface $c) {
            $db1Settings = $c->get('settings')['db1'];
            $dsn = 'mysql:host=' . $db1Settings['host'] . ';dbname=' . $db1Settings['dbname'];
            $options = [ ... ];
            return new PDO($dsn, $db1Settings['user'], $db1Settings['pass'],$options);
        },
        'db2' => function (ContainerInterface $c) {
            $db2Settings = $c->get('settings')['db2'];
            $dsn = 'mysql:host=' . $db2Settings['host'] . ';dbname=' . $db2Settings['dbname'];
            $options = [ ... ];
            return new PDO($dsn, $db2Settings['user'], $db2Settings['pass'],$options);
        },

    ]);
};

但這是最好的方法嗎?以及如何在不必注入整個容器的情況下訪問存儲庫類中的連接?

But is it the best way to do it? And how can I access the connections in a repository class without having to inject the whole container?

推薦答案

您有多種選擇:

1. 擴展 PDO
2. 自動裝配的對象

1.擴展 PDO

use PDO;

class PDO2 extends PDO
{
    // must be empty
}

容器定義:

use PDO2;

// ...

return [
    PDO::class => function (ContainerInterface $container) {
        return new PDO(...);
    },

    PDO2::class => function (ContainerInterface $container) {
        return new PDO2(...);
    },
];

使用

use PDO;
use PDO2;

class MyRepository
{
    private $pdo;

    private $pdo2;
    
    public function __construct(PDO $pdo, PDO2 $pdo2)
    {
        $this->pdo = $pdo;
        $this->pdo2 = $pdo2;
    }
}

2.自動裝配對象

參見 Matthieu Napoli 的回答:https://stackoverflow.com/a/57758106/1461181

See Matthieu Napoli's answer: https://stackoverflow.com/a/57758106/1461181

這篇關于如何在 slim 4 中設置和注入多個 PDO 數據庫連接?的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！