Php Symfony原则多对多插入
我在Symfony中的实体和控制器有问题。我会在生成的多对多表中插入DB值 仅包含多对多元素的实体请求Php Symfony原则多对多插入,php,entity-framework,symfony,doctrine-orm,doctrine,Php,Entity Framework,Symfony,Doctrine Orm,Doctrine,我在Symfony中的实体和控制器有问题。我会在生成的多对多表中插入DB值 仅包含多对多元素的实体请求 class Requests { /** * @ORM\ManyToMany(targetEntity="Tipi", inversedBy="requests") * @ORM\JoinTable(name="tipi_richieste") */ private $tipi; public function __construct() { $this->ti
class Requests {
/**
* @ORM\ManyToMany(targetEntity="Tipi", inversedBy="requests")
* @ORM\JoinTable(name="tipi_richieste")
*/
private $tipi;
public function __construct() {
$this->tipi = new \Doctrine\Common\Collections\ArrayCollection();
}
/**
* Add tipi
*
* @param \AppBundle\Entity\Tipi $tipi
*
* @return Requests
*/
public function addTipi(\AppBundle\Entity\Tipi $tipi) {
$this->tipi[] = $tipi;
return $this;
}
/**
* Remove tipi
*
* @param \AppBundle\Entity\Tipi $tipi
*/
public function removeTipi(\AppBundle\Entity\Tipi $tipi) {
$this->tipi->removeElement($tipi);
}
/**
* Get tipi
*
* @return \Doctrine\Common\Collections\Collection
*/
public function getTipi() {
return $this->tipi;
}
}
class Tipi {
/**
* @ORM\ManyToMany(targetEntity="Requests", mappedBy="tipi")
*/
private $requests;
/**
* Constructor
*/
public function __construct() {
$this->requests = new \Doctrine\Common\Collections\ArrayCollection();
}
/**
* Add request
*
* @param \AppBundle\Entity\Requests $request
*
* @return Tipi
*/
public function addRequest(\AppBundle\Entity\Requests $request)
{
$this->requests[] = $request;
return $this;
}
/**
* Remove request
*
* @param \AppBundle\Entity\Requests $request
*/
public function removeRequest(\AppBundle\Entity\Requests $request)
{
$this->requests->removeElement($request);
}
/**
* Get requests
*
* @return \Doctrine\Common\Collections\Collection
*/
public function getRequests()
{
return $this->requests;
}
}
仅包含多对多元素的实体Tipi
class Requests {
/**
* @ORM\ManyToMany(targetEntity="Tipi", inversedBy="requests")
* @ORM\JoinTable(name="tipi_richieste")
*/
private $tipi;
public function __construct() {
$this->tipi = new \Doctrine\Common\Collections\ArrayCollection();
}
/**
* Add tipi
*
* @param \AppBundle\Entity\Tipi $tipi
*
* @return Requests
*/
public function addTipi(\AppBundle\Entity\Tipi $tipi) {
$this->tipi[] = $tipi;
return $this;
}
/**
* Remove tipi
*
* @param \AppBundle\Entity\Tipi $tipi
*/
public function removeTipi(\AppBundle\Entity\Tipi $tipi) {
$this->tipi->removeElement($tipi);
}
/**
* Get tipi
*
* @return \Doctrine\Common\Collections\Collection
*/
public function getTipi() {
return $this->tipi;
}
}
class Tipi {
/**
* @ORM\ManyToMany(targetEntity="Requests", mappedBy="tipi")
*/
private $requests;
/**
* Constructor
*/
public function __construct() {
$this->requests = new \Doctrine\Common\Collections\ArrayCollection();
}
/**
* Add request
*
* @param \AppBundle\Entity\Requests $request
*
* @return Tipi
*/
public function addRequest(\AppBundle\Entity\Requests $request)
{
$this->requests[] = $request;
return $this;
}
/**
* Remove request
*
* @param \AppBundle\Entity\Requests $request
*/
public function removeRequest(\AppBundle\Entity\Requests $request)
{
$this->requests->removeElement($request);
}
/**
* Get requests
*
* @return \Doctrine\Common\Collections\Collection
*/
public function getRequests()
{
return $this->requests;
}
}
insert的表单类型是EntityType
->add('tipi', EntityType::class, array(
'label' => 'Tipo',
'class' => 'AppBundle:Tipi',
'mapped' => false,
'attr' => array('class' => 'form-control'),
'multiple' => true,
'by_reference' => false,
'query_builder' => function (EntityRepository $er) {
return $er->createQueryBuilder('t');
},
))
在我的控制器中,我是这样工作的:
public function indexAction(Request $request) {
$requests = new Requests();
$em = $this->getDoctrine()->getManager();
$form = $this->createForm(RequestsType::class, $requests);
$form->handleRequest($request);
if ($form->isSubmitted() && $form->isValid()) {
$requests->setCreateAt(new \DateTime('now'));
$request_tipi = $form["tipi"]->getData();
$tipi_array = [];
die($form["tipi"]->getData());
$tipi_array = $em->getRepository('AppBundle:Tipi')->findOneBy(array('codice' => $form["tipi"]->getData()));
$tipi = new Tipi();
$requests->addTipi($form["tipi"]->getData());
$em->persist($requests);
$em->flush();
//return $this->redirectToRoute('immovable_edit', array('id' => $immovables->getId()));
}
return $this->render('AppBundle:Requests:index.html.twig', array(
'requests' => $requests,
'form' => $form->createView(),
));
}
当我为返回$form[“tipi”]->getData()的值设置die时,我得到一个数组集合:
Doctrine\Common\Collections\ArrayCollection@000000005b52ae6b00000000731dd0b4
但是我得到了这个错误
可捕获的致命错误:传递给AppBundle\Entity\Requests::addTipi()的参数1必须是AppBundle\Entity\Tipi的实例,Doctrine\Common\Collections\ArrayCollection的实例,在第31行的C:\xampp\htdocs\bugaro\src\AppBundle\Controller\RequestsController.php中调用并定义
Request::addTipi()
fuction将singletipi添加到其内部集合中。因此,您不能在调用中添加整个ArrayCollection
你有两个选择
为每个人做一个foreach
foreach($form["tipi"]->getData() as $tipi) {
$requests->addTipi($tipi);
}
public function addTipis($tipis) {
foreach($tipis as $tipi) {
$this->tipi[] = $tipi;
}
return $this;
}
创建类似于请求的多字母::addCollections(ArrayCollection[])
foreach($form["tipi"]->getData() as $tipi) {
$requests->addTipi($tipi);
}
public function addTipis($tipis) {
foreach($tipis as $tipi) {
$this->tipi[] = $tipi;
}
return $this;
}
无论如何,你的代码中有很多东西需要修正。这里的一个帖子太多了
但这些修复中最重要的是,您不需要大部分控制器代码。;-)
由于您将$requests
对象传递到表单中,它已经是双向组合的,这意味着Symfony的表单组件应该自动用新值填充它的属性。这包括多对多关系的集合
另外,如果您不传递该对象,$form->getData()
应该返回一个新的Resnponses
实例,因此您不需要创建它并手动传递,直到它成为现有实例的版本。请求::addTipi()
函数将singletipi添加到其内部集合。因此,您不能在调用中添加整个ArrayCollection
你有两个选择
为每个人做一个foreach
foreach($form["tipi"]->getData() as $tipi) {
$requests->addTipi($tipi);
}
public function addTipis($tipis) {
foreach($tipis as $tipi) {
$this->tipi[] = $tipi;
}
return $this;
}
创建类似于请求的多字母::addCollections(ArrayCollection[])
foreach($form["tipi"]->getData() as $tipi) {
$requests->addTipi($tipi);
}
public function addTipis($tipis) {
foreach($tipis as $tipi) {
$this->tipi[] = $tipi;
}
return $this;
}
无论如何,你的代码中有很多东西需要修正。这里的一个帖子太多了
但这些修复中最重要的是,您不需要大部分控制器代码。;-)
由于您将$requests
对象传递到表单中,它已经是双向组合的,这意味着Symfony的表单组件应该自动用新值填充它的属性。这包括多对多关系的集合
另外,如果不传递该对象,$form->getData()
应该返回一个新的Resnponses
实例,因此,您不需要创建它并手动传递,直到它成为现有实例的版本。我个人发现EntityType
最适合一对多和多对多-您已经尝试过了吗?@kingkero是,它很好用。我个人发现EntityType
最适合一对多和多对多-你已经试过了吗?@kingkero是的,它很好用。嗨,对不起。我用collectiontype实现它,现在我收到关联字段“AppBundle\Entity\Requests#$tipi”的类型为“条令\Common\Collections\Collection |数组”的错误预期值,改为“数组”。。。嗨,对不起。我用collectiontype实现它,现在我收到关联字段“AppBundle\Entity\Requests#$tipi”的类型为“条令\Common\Collections\Collection |数组”的错误预期值,改为“数组”。。。