我有兩個表backup_policy
和backup_settings
。 backup_policy
有一個名爲backup_settings_id
的外鍵字段,映射到backup_settings
id
字段。這是執行以創建兩個表Symfony3外鍵關係實體不工作
ALTER TABLE `backup_policy_scheduler`
ADD FOREIGN KEY (`backup_policy_id`) REFERENCES `backup_policy` (`id`)
現在的問題是,當我嘗試這個代碼,屏幕上不斷加載,給我最大的執行誤差之間的關係查詢。
$repository=$this->getDoctrine()>getRepository('AppBundle:BackupPolicy');
$settings = $repository->findAll();
當我準備一個sql語句,並執行fetchAll()它獲取結果。
當我嘗試插入記錄它給了我下面的錯誤:
Type error: Argument 1 passed to AppBundle\Entity\BackupPolicy::setBackupSettings() must be an instance of AppBundle\Entity\BackupSettings, string given, called in /var/www/symfonyproject/src/AppBundle/Controller/BackupPolicyController.php on line 46
<?php
namespace AppBundle\Entity;
use Doctrine\ORM\Mapping as ORM;
/**
* BackupPolicy
*
* @ORM\Table(name="backup_policy")
*@ORM\Entity(repositoryClass="AppBundle\Entity\Repository\BackupPolicyRepository")
*/
class BackupPolicy
{
/**
* @var integer
*
* @ORM\Column(name="id", type="integer", nullable=false)
* @ORM\Id
* @ORM\GeneratedValue(strategy="IDENTITY")
*/
private $id;
/**
* @var \BackupSettings
*
* @ORM\ManyToOne(targetEntity="BackupSettings")
* @ORM\JoinColumns({
* @ORM\JoinColumn(name="backup_settings_id", referencedColumnName="id")
* })
*/
private $backupSettings;
/**
* Get id
*
* @return integer
*/
public function getId()
{
return $this->id;
}
/**
* Set backupSettings
*
* @param \AppBundle\Entity\BackupSettings $backupSettings
*
* @return BackupPolicy
*/
public function setBackupSettings(\AppBundle\Entity\BackupSettings $backupSettings = null)
{
$this->backupSettings = $backupSettings;
return $this;
}
/**
* Get backupSettings
*
* @return \AppBundle\Entity\BackupSettings
*/
public function getBackupSettings()
{
return $this->backupSettings;
}
}
你的錯誤是,你嘗試分配$這個 - > backupSettings一個字符串。在你的控制器行46中。這與foreignn密鑰或映射無關。 顯示您的控制器。 – Alsatian
是的,我說得對。我只插入一個字符串。當我插入backup_settings表時,我得到了這個。但爲什麼即使findAll()不工作? –
請顯示控制器。 – Alsatian