php – Symfony2.1映射错误:class_parents()
作者:互联网
我试图在Symfony2.1项目中使用Doctrine2从表(通过实体)获取数据时遇到问题.这是我得到错误的控制器:
/**
* Country list
*/
public function countrylistAction()
{
$em = $this->getDoctrine()->getManager();
$countryList = $em->getRepository('ProjectBaseBundle:SYS_TCountry')
->findAll();
$serializer = new Serializer(array(new GetSetMethodNormalizer()),
array('json' => new JsonEncoder()));
return new Response($serializer->serialize($countryList, 'json'));
}
实体:
<?php
namespace Company\Project\BaseBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
use Symfony\Component\Validator\Constraints as Assert;
/**
* @ORM\Entity
* @ORM\Table(name="SYS_TCountry")
*/
class SYS_TCountry
{
/**
* @ORM\Id
* @ORM\Column(type="string", length=3, nullable=false)
* @var string
*/
protected $idcountry;
/**
* @ORM\Column(type="string", length=75, nullable=false)
* @Assert\NotBlank()
* @var string
*/
protected $name;
....
public function getIdcountry() { return $this->idcountry; }
public function getName() { return $this->name; }
public function getA2() { return $this->a2; }
public function getA3() { return $this->a3; }
public function getIdstatus() { return $this->idstatus; }
public function setIdcountry($idcountry) { $this->idcountry = $idcountry; }
public function setName($name) { $this->name = $name; }
public function setA2($a2) { $this->a2 = $a2; }
public function setA3($a3) { $this->a3 = $a3; }
public function setIdstatus($idstatus) { $this->idstatus = $idstatus; }
public function __toString() { return $this->idcountry; }
}
Config.yml:
# Doctrine Configuration
doctrine:
dbal:
driver: %database_driver%
host: %database_host%
port: %database_port%
dbname: %database_name%
user: %database_user%
password: %database_password%
charset: UTF8
orm:
auto_generate_proxy_classes: %kernel.debug%
auto_mapping: true
这是错误:
Warning: class_parents():
Class Company\Project\BaseBundle\Entity\SYS_TCountry does not exist and could not be loaded in
/var/www/project/src/vendor/doctrine/common/lib/Doctrine/Common/Persistence/Mapping/RuntimeReflectionService.php line 40
这很奇怪,因为正如Doctrine在控制台中说的那样,映射已经正确完成了:我测试它执行php app / console doctrine:mapping:info:
[OK] Company\Project\BaseBundle\Entity\SYS_TCountry
如果我在控制台中执行查询一切顺利 – > app / console doctrine:query:sql’SELECT * FROM SYS_TCountry’,返回结果.
我不知道是否使用Symfony2.1我必须配置与2.0版本不同的东西,但似乎是相同的,因为映射是Doctrine的责任.
解决方法:
Symfony遵循PSR-0标准的文件名.除其他外,这意味着如果你在类名中使用下划线,它将在决定你的类应该存在的位置时用目录分隔符替换它,如下所示:
\namespace\package\Class_Name => /path/to/project/lib/vendor/namespace/package/Class/Name.php
所以,如果你有一个名为SYS_TCountry的类,它会期望找到它
Company/Project/BaseBundle/Entity/SYS/TCountry.php
代替
Company/Project/BaseBundle/Entity/SYS_TCountry.php
我认为您最好的解决方案是将文件名和类名更改为SYSTCountry.您无需更改表名称.
标签:php,orm,doctrine,symfony-2-1 来源: https://codeday.me/bug/20190530/1182297.html