无法对工作流程“未命名”应用转换“拒绝”

Question

我在新的symfony（3.2）工作流功能方面遇到了一些麻烦。我似乎只能将“审核”过渡应用于任务。

$workflow->apply($task, 'review');

看这段代码有用。

拒绝和审查我总是遇到这个错误。

这里是控制器和实体：

控制器

namespace AppBundle\Controller;

use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\Workflow\DefinitionBuilder;
use Symfony\Component\Workflow\MarkingStore\SingleStateMarkingStore;
use Symfony\Component\Workflow\Registry;
use Symfony\Component\Workflow\Transition;
use Symfony\Component\Workflow\Workflow;

class WorkflowController extends Controller
{
    /** @Route("/workflow", name="workflow") */
    public function indexAction(
        Request $request
    ) {
        $registry = new Registry();

        $definition = (new DefinitionBuilder())
            ->addPlaces(['drafted', 'reviewed', 'rejected', 'published'])
            ->addTransition(new Transition('review', 'drafted', 'reviewed'))
            ->addTransition(new Transition('publish', 'reviewed', 'published'))
            ->addTransition(new Transition('reject', 'reviewed', 'rejected'))
            ->build();

        $workflow = new Workflow(
            $definition,
            new SingleStateMarkingStore('state')
        );

        $task = new \AppBundle\Entity\Task();

        $workflow->apply($task, 'reject');

        $registry->add($workflow, \AppBundle\Entity\Task::class);

        return new Response(json_encode([
            'available_actions' => [
                'review'  => $workflow->can($task, 'review'),
                'publish' => $workflow->can($task, 'publish'),
                'reject'  => $workflow->can($task, 'reject'),
            ]]
        ));
    }
}

实体

namespace AppBundle\Entity;

use Doctrine\ORM\Mapping as ORM;

class Task
{
    private $id;

    private $title;

    public $state;

    public function getId()
    {
        return $this->id;
    }

    public function setTitle($title)
    {
        $this->title = $title;

        return $this;
    }

    public function getTitle()
    {
        return $this->title;
    }
}

有人可以帮我解决问题吗？

Answer 1

是的，您只能对新创建的任务应用review，因为它们处于第一个状态drafted。要发布或拒绝任务，您首先需要应用review，然后reject或publish：

$task = new Task();

// Place newly created task to reviewed status
$workflow->apply($task, 'reject');

// Then reject or publish
$workflow->apply($task, 'publish')

我会尽力解释一下。首先，您可以定义任务的状态：

->addPlaces(['drafted', 'reviewed', 'rejected', 'published'])

新任务获得drafted（列表中的第一个）状态，因为您省略了Task s state属性的设置值。

接下来，您定义任务从一个状态传递到另一个状态的规则（称为Transition）

->addTransition(new Transition('review', 'drafted', 'reviewed'))

它刚刚定义了名为review的转换（或操作），仅适用于drafted任务，在执行任务后，其状态更改为reviewed。

因此，总而言之，首先应审查所有新创建的任务，然后发布或拒绝。