How do inject array of implementations into constructor in PHP

Question

How do inject array of implementations into class through constructor. I am sharing the link which is c#. I want to achieve the same in php.

How to achieve same in php.

public interface IFoo { }
public class FooA : IFoo {}
public class FooB : IFoo {}

public class Bar
{
    //array injected will contain [ FooA, FooB ] 
    public Bar(IFoo[] foos) { }
}

public class MyModule : NinjectModule
{
    public override void Load()
    {
        Bind<IFoo>().To<FooA>();
        Bind<IFoo>().To<FooB>();
        //etc..
    }
}

https://stackoverflow.com/a/13383476/1844634

Thanks in advance.

112Legion · Answer 1 · 2020-02-03T17:39:42.783

1

PHP does not support generics duo to run-time performance difficulty. So there are no way to explain that you expect all interfaces via definition of contractor. So you have to configure DI container manually. To explicitly tell that your class needs all classes that support some kind of interface.

Laravel for configuration use ServiceProvider to do all kind of configuration: In the class \App\Providers\AppServiceProvider you can configure creation of your class.


    public function register(): void
    {
        // to configure implementation for an interface or abstract class
        // you can only configure one implementation for interface
        $this->app->bind(\App\IFoo::class, \App\FooA::class);

        // or 'tag' several implementation for one string tag.
        $this->app->tag([\App\FooA::class, \App\FooB::class], \App\IFoo::class);

        $this->app->bind(\App\Bar::class, function(\Illuminate\Contracts\Foundation\Application $container){
            // get all tagged implementations
            $foos = $container->tagged(\App\IFoo::class);

            return new \App\Bar($foos);
        });
    }

edited Feb 03 '20 at 17:39

answered Feb 03 '20 at 14:25

112Legion

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How i can do that? Any reference? – user1844634 Feb 03 '20 at 14:26
@user1844634 sure. I will describe how to configure DI container. – 112Legion Feb 03 '20 at 14:27
There is no need to interfaces here? Directly we are dealing with the concrete classes . How to bind the interface to concrete classes? – user1844634 Feb 03 '20 at 14:40
@user1844634 it's quite simile will add to my answer. – 112Legion Feb 03 '20 at 14:42
Please take the above example that i mentioned in the question and write your answer, It will help me alot. Thanks in advance. – user1844634 Feb 03 '20 at 14:44
@user1844634 I have used the same naming like in your questions. PS: To all classes I added. 'App' namespace cos you will never put your class into global namespace. – 112Legion Feb 03 '20 at 15:00
Why this class is needed \App\YourClass::class? – user1844634 Feb 03 '20 at 17:31
@user1844634 sorry for the mistake it should be Bar class according to your question – 112Legion Feb 03 '20 at 17:39
when i used the bar class , it's throwing exception as Undefined constant. – user1844634 Feb 03 '20 at 18:01
Let us [continue this discussion in chat](https://chat.stackoverflow.com/rooms/207119/discussion-between-user1844634-and-112legion). – user1844634 Feb 03 '20 at 18:01

Foued MOUSSI · Answer 2 · 2020-02-03T14:36:07.930

You may need to use Tagging. For example, perhaps you are building a report aggregator that receives an array of many different Report interface implementations. After registering the Report implementations, you can assign them a tag using the tag method:

$this->app->bind('App\Reports\MemoryReportInterface', 'App\Reports\MemoryReportImplementation');       
$this->app->bind('App\Reports\SpeedReportInterface', 'App\Reports\SpeedReportImplementation');  

$this->app->tag(['App\Reports\MemoryReportInterface', 'App\Reports\MemoryReportInterface'], 'reports');

Once the services have been tagged, you may easily resolve them all via the tagged method:

$this->app->bind('ReportAggregator', function ($app) {
    return new ReportAggregator($app->tagged('reports'));
});

Usage

<?php 

namespace ...;

/**
 * 
 */
class ReportAggregator
{
    private $reports;

    function __construct($reports)
    {
        $this->reports = $reports;
    }

    public function getReports() {
        return $this->reports;
    }
    //...
}

I want my class to depend on abstractions.Rather than the concrete implementations. How i can achieve here? — user1844634, Feb 03 '20 at 14:34
If i am not wrong it should be same interface. Same interface implemented by two concrete classes. It should be either of MemoryReportInterface or SpeedReportInterface — user1844634, Feb 03 '20 at 14:48
please look the example at that i mentioned in the question and update the answer please. Thanks in advance. — user1844634, Feb 03 '20 at 14:52

How do inject array of implementations into constructor in PHP

2 Answers2