Symfony2服务容器 - 将普通参数传递给服务构造函数

Question

我有这个Paginator类构造函数:

class Paginator
{    
    public function __construct($total_count, $per_page, $current_page)
    {
    }
}

该Paginator服务注册Ibw/JobeetBundle/Resources/config/services.yml如下:

parameters:
    ibw_jobeet_paginator.class: Ibw\JobeetBundle\Utils\Paginator

services:
    ibw_jobeet_paginator:
        class: %ibw_jobeet_paginator.class%

当我使用Paginator这样的时候:

$em = $this->getDoctrine()->getManager();

$total_jobs = $em->getRepository('IbwJobeetBundle:Job')->getJobsCount($id);
$per_page = $this->container->getParameter('max_jobs_on_category');
$current_page = $page; 

$paginator = $this->get('ibw_jobeet_paginator')->call($total_jobs, $per_page, $current_page);

我得到这个例外:

警告:缺少Ibw\JobeetBundle\Utils\Paginator :: __ construct()的参数1,在1306行的/var/www/jobeet/app/cache/dev/appDevDebugProjectContainer.php中调用,并在/ var/www/jobeet /中定义src/Ibw/JobeetBundle/Utils/Paginator.php第13行

我猜在将参数传递给Paginator服务构造函数时出现了问题.你能告诉我,如何将参数传递给服务构造函数？

Answer 1

好吧,要回答你的问题,你使用arguments参数传递服务构造函数参数:

services:
    ibw_jobeet_paginator:
        class: %ibw_jobeet_paginator.class%
    arguments:
        - 1 # total
        - 2 # per page
        - 3 # current page

当然,由于参数是动态的,因此这并没有真正帮助你.

而是将参数从构造函数移动到另一个方法:

class Paginator
{    
    public function __construct() {}

    public function init($total_count, $per_page, $current_page)
    {
    }
}

$paginator = $this->get('ibw_jobeet_paginator')->init($total_jobs, $per_page, $current_page);