symfony - 当我尝试创建新的时,php bin/console make:crud 中有错误
问题描述
我在控制台中创建了一个 crud
php bin/console make:crud Location
它工作正常。我去/location/
单击新建,我收到此 sql 错误
执行 'SELECT e0_.id AS id_0, e0_.created AS created_1, e0_.updated AS updated_2, e0_.updated_by_id AS updated_by_id_3, e0_.type AS type_4, p1_.name AS name_5, p1_.email AS email_6, e0_ 时发生异常.created_by_id AS created_by_id_7, e0_.permission_id AS permission_id_8, p1_.relationships_id AS relationship_id_9, p1_.locations_id AS locations_id_10 FROM people p1_':SQLSTATE [42S22]:未找到列:1054 '字段中的未知列'e0_.id'
好像那里有两张桌子。请帮忙。
下面是location
课堂。它包括people
在其中。创建者是一个person
. 我不知道我是否有正确的映射,我很新Symfony
。
namespace App\Entity;
use Doctrine\ORM\Mapping as ORM;
/**
* Class Location
* @package App\Entity
* @ORM\Entity()
* @ORM\Table("locations")
*/
class Location
{
/**
* @ORM\Id
* @ORM\GeneratedValue
* @ORM\Column(type="integer")
*/
protected $id;
/**
* @ORM\Column(type="datetime")
*/
protected $created;
/**
* @ORM\Column(type="datetime")
*/
protected $updated;
/**
* @ORM\OneToOne(targetEntity="App\Entity\Person")
* @ORM\JoinColumn(name="created_by_id", referencedColumnName="id")
*/
protected $createdBy;
/**
* @ORM\OneToOne(targetEntity="App\Entity\Permission")
* @ORM\JoinColumn(name="permission_id", referencedColumnName="id")
*/
protected $permissions;
/**
* @ORM\Column(name="updated_by_id")
* @ORM\OneToOne(targetEntity="App\Entity\Person")
*/
protected $updatedBy;
/**
* @ORM\Column(length=255)
*/
private $name;
/**
* @ORM\OneToOne(targetEntity="Address")
* @ORM\JoinColumn(name="address_id", referencedColumnName="id")
*/
private $address;
/**
* @ORM\Column(length=255)
*/
private $geoSpatial;
}
解决方案
/**
* @ORM\ManyToOne(targetEntity="App\Entity\Person")
* @ORM\JoinColumn(name="created_by_id", referencedColumnName="id")
*/
protected $createdBy;
创建者需要正确映射。我真是个傻逼!
感谢那些注意到“人们从哪里来”帮助我的人,有时您在自己的代码中看不到东西!
推荐阅读
- python - 重复的字符串清理,如何使其更简洁
- java - Java中ntlm的svnkit身份验证问题
- flutter - 颤振错误图像:无法加载资产
- java - 如何正确地从 Spring WebFlux 中的多个 Flux (WebsocketSession::receive) 向 Sink 发出值?
- node.js - Telegram Bot 可以私聊吗?
- c# - Blazor 在其他组件内设置值
- python - 如何配置 Codacy Python 短绒?
- c++ - 通过函数移动 QGraphicsPixmapItem
- typescript - 重载函数的类型推断
- node.js - `BelongsTo` 关系有什么好处?