Question

我有两个实体：门票和类别

class tickets 
{
    /**
     * @var integer $id
     *
     * @ORM\Column(name="id", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    private $id;
...
    /**
     * @ORM\OneToOne(targetEntity="categories", mappedBy="ticket")
     */
    private $categories;

public function setCategories($categories)
    {
        $this->categories= $categories;
    }

    public function getCategories()
    {
        return $this->categories;
    }
}

class categories
{
    /**
     * @var integer $id
     *
     * @ORM\Column(name="id", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    private $id;

    /**
     * @var integer $id_ticket
     *
     * @ORM\Column(name="id_ticket", type="integer")
     */
    private $id_ticket;
...
    /**
     * @ORM\OneToOne(targetEntity="tickets", inversedBy="categories")
     * @ORM\JoinColumn(name="id", referencedColumnName="id_ticket")
     */
    private $ticket;

public function setTicket()
    {
        $this->ticket= $ticket;

    }
public function getTicket()
    {
        return $this->ticket;
    }

在我的门票控制器中创建新票证时，我想创建一个新类别，其中tickets.id = categories.id_ticket

$ticket= new tickets();
$category= new categories();
$ticket->setCategories($categories);
$em->persist($categories);
$em->persist($ticket);
$em->flush();

我收到此错误：

SQLSTATE [23000]：[Microsoft] [SQL Server Native Client 10.0] [SQL Server]当IDENTITY_INSERT设置为OFF时，无法在表'categories'中为identity列插入显式值。

它尝试做的插入是：

INSERT INTO categories(id_ticket, ... , **id**) VALUES (?, ..., **?**)
Parameters: { 1: 'null', ..., **18: 'null'** }

我做错了什么？我不知道为什么它试图插入“id”。我知道这就是我收到错误的原因，但我是symfony2和doctrine的新手，我不知道如何解决它。

Answer 1

您的拥有方存在问题（查看here）

尝试：

$ticket= new tickets();
$category= new categories();
$categories->setTicket($ticket);
$em->persist($ticket);
$em->persist($categories);
$em->flush();

OneToOne关系和设置实体

1 个答案: