我有两个名为Question和SolutionMethod的PHP模型类。类别可能包含许多项目,而项目可能属于许多类别。我已经为这两个类创建了ManyToMany关系:
class Question {
/**
* @ORM\ManyToMany(targetEntity="SolutionMethod", mappedBy="questions", cascade={"persist"})
*/
private $solutions;
/**
* Add items
*
* @param Acme\MyBundle\Entity\SolutionMethod $solution
*/
public function addItems(\Acme\MyBundle\Entity\SolutionMethod $solution) {
$this->solutions[] = $solutions;
}
/**
* Get solutions
*
* @return Doctrine\Common\Collections\Collection
*/
public function getSolutions() {
return $this->solutions;
}
[...]
}
和
class SolutionMethod {
/**
* @ORM\ManyToMany(targetEntity="Question", inversedBy="solutions", cascade={"persist"})
* @ORM\JoinTable(name="question_solution")
*/
private $questions;
/**
* Add questions
*
* @param Acme\MyBundle\Entity\Question $question
*/
public function addCategories(\Acme\MyBundle\Entity\Question $question) {
$this->questions[] = $question;
}
/**
* Get questions
*
* @return Doctrine\Common\Collections\Collection
*/
public function getQuestions() {
return $this->questions;
}
[...]
}
在我的表单中,此关联使用带有必需属性的多选项进行呈现,但我有一些没有SolutionMethod的问题,但我无法保存,并且显示所需的消息。
解决方案很简单,让我觉得自己像个白痴
class QuestionType extends AbstractType {
public function buildForm(FormBuilder $builder, array $options) {
$builder
->add('solutions', null, array('required' => false))
[...]
;
}
}