Symfony2服务容器 - 将普通参数传递给服务构造函数

时间:2013-10-02 17:13:48

标签: php symfony constructor dependency-injection pagination

我有这个Paginator类构造函数:

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

Paginator服务已在Ibw/JobeetBundle/Resources/config/services.yml中注册,如下所示:

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

services:
    ibw_jobeet_paginator:
        class: %ibw_jobeet_paginator.class%

当我像这样使用Paginator时:

$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);

我得到了这个例外:

  

警告:缺少参数1   Ibw \ JobeetBundle \ Utils \ Paginator :: __ construct(),调用   /var/www/jobeet/app/cache/dev/appDevDebugProjectContainer.php上线   1306并定义于   /var/www/jobeet/src/Ibw/JobeetBundle/Utils/Paginator.php第13行

我认为将参数传递给Paginator服务构造函数有问题。你能告诉我,如何将参数传递给服务构造函数?

1 个答案:

答案 0 :(得分:23)

那么,要回答您的问题,请使用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,因此并没有真正帮助你。

相反,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);