User.php;
user.php的;
<?php
namespace Acme\DemoBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
/**
* @ORM\Entity
* @ORM\Table(name="like")
* @ORM\InheritanceType("SINGLE_TABLE")
* @ORM\DiscriminatorColumn(name="itemtype", type="integer")
* @ORM\DiscriminatorMap({
* "1" = "LikeArticle",
* "2" = "LikePage",
* "3" = "LikeSite",
* })
*/
class Like
{
/**
* @ORM\Column(type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
protected $id;
/**
* @ORM\Column(type="integer")
*/
protected $userid;
/**
* @ORM\ManyToOne(targetEntity="User", inversedBy="likes")
* @ORM\JoinColumn(name="userid", referencedColumnName="id")
*/
protected $user;
}
LikePage.php;
LikePage.php;
<?php
namespace Acme\DemoBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
/**
* @ORM\Entity
* @ORM\Table(name="like_page")
*/
class LikePage extends Like
{
/**
* @ORM\Column(type="integer")
*/
protected $itemid;
/**
* @ORM\ManyToOne(targetEntity="Page", inversedBy="likes")
* @ORM\JoinColumn(name="itemid", referencedColumnName="id")
*/
protected $page;
LikeArticle.php;
LikeArticle.php;
namespace Acme\DemoBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
/**
* @ORM\Entity
* @ORM\Table(name="like_article")
*/
class LikePage extends Like
{
/**
* @ORM\Column(type="integer")
*/
protected $itemid;
/**
* @ORM\ManyToOne(targetEntity="Article", inversedBy="likes")
* @ORM\JoinColumn(name="itemid", referencedColumnName="id")
*/
protected $article;
I add data to the database in this way. I'm trying to retrieve data from the Page Entity. But, When executing this;
我以这种方式向数据库添加数据。我正在尝试从Page Entity中检索数据。但是,执行此时;
$likes = $page->getLikes();
foreach($likes as $like) {
}
I am getting this error;
我收到了这个错误;
An exception occurred while executing 'SELECT t0.id AS id1, t0.userid AS userid2, t0.itemid AS itemid3, t0.userid AS userid4, t0.itemid AS itemid5, t0.itemtype FROM like t0 WHERE t0.itemid = ? AND t0.itemtype IN ('1')' with params [1]:
SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'like t0 WHERE t0.itemid = '1' AND t0.itemtype IN ('1')' at line 1
In this case what should I do? Why not find itemid?
在这种情况下,我该怎么办?为什么找不到itemid?
1 个解决方案
#1
13
like is a reserved word,escape it or better yet change your table name.
喜欢是一个保留字,逃避它或更好地改变你的表名。
文件
Change it to
将其更改为
* @ORM\Table(name="`like`")
#1
13
like is a reserved word,escape it or better yet change your table name.
喜欢是一个保留字,逃避它或更好地改变你的表名。
文件
Change it to
将其更改为
* @ORM\Table(name="`like`")