symfony教义协会 [英] symfony doctrine associations

问题描述

所以我设法让这个工作与用户和工作之间的关系

我遵循完全相同的过程，检查文档，我失去了

我正在设置的关系是工作工作

1分配了许多工作项的作业

../ Entity / Job.php

  ** 
 * @ ORM\OneToMany（targetEntity =Work，mappedBy =workJob）
 * 
 * / 
 protected $ jobToWork; 
 / ** 
 *添加jobToWork 
 * 
 * @param \Laelaps\InvoiceBundle\Entity\Work $ jobToWork 
 * @return作业
 * / 
 public function addJobToWork（\Laelaps\InvoiceBundle\Entity\Work $ jobToWork）
 {
 
 $ this-> jobToWork [] = $ jobToWork; 
 
 return $ this; 
} 
 / ** 
 *删除jobToWork 
 * 
 * @param \Laelaps\InvoiceBundle\Entity\Work $ jobToWork 
 * / 
 public function removeJobToWork（\Laelaps\InvoiceBundle\Entity\Work $ jobToWork）
 {
 $ this-> jobToWork-> removeElement（$ jobToWork）; 
} 
 
 / ** 
 *获取jobToWork 
 * 
 * @return \Doctrine\Common\Collections\Collection 
 * / 
 public function getJobToWork（）
 {
 return $ this-> jobToWork; 
} 
  

../ Entity / Work.php

  / ** 
 * @ ORM\ManyToOne（targetEntity =Job，inversedBy =jobToWork）
 * @ ORM\\ \\ JoinColumn（name =work_job，referencedColumnName =id）
 * / 
 protected $ workJob; 
 
 / ** 
 *设置workJob 
 * 
 * @param integer $ workJob 
 * @return工作
 * / 
 public function setWorkJob（$ workJob）
 {
 $ workJob-> addJobToWork（$ this）; 
 $ this-> workJob = $ workJob; 
 
 return $ this; 
} 
 
 / ** 
 * Get workJob 
 * 
 * @return integer 
 * / 
 public function getWorkJob ）
 {
 return $ this-> workJob; 
} 
  

我做了一个应用程序/控制台原则：generate：crud

我知道为我的用户工作的工作关系设置完全一样。

当我尝试去创建一个新的工作项目时，我得到的错误是
类.. \Entity\Job的对象不能转换为字符串

任何帮助，一如既往的非常赞赏

更新 / p>

如答案中所定义

  public function __toString（）
 {
 return（string）$ this-> getJobName（）; 
} 
  

当我使用FOSUserBundle和我的用户类扩展FOSUser类时，这意味着__toString（）函数在FOSUser中定义，因此在向用户创建关联时没有问题。

解决方案

您可能会尝试将作业显示为字符串（例如在选择框或列表中）。只需向Job实体添加一个__toString（）方法，如下所示：

  public function __toString（）
 {
 return $ this-> name; //例如
} 
  

Hi so I managed to get this working for a relationship between users and a Job

I followed exactly the same procedure, checked the documentation and I am at a loss

The relationship i'm trying to set up is Job to work

1 Job with many work items assigned to it

../Entity/Job.php

 /**
 * @ORM\OneToMany(targetEntity="Work",mappedBy="workJob")
 *
 */
protected $jobToWork;
/**
 * Add jobToWork
 *
 * @param \Laelaps\InvoiceBundle\Entity\Work $jobToWork
 * @return Job
 */
public function addJobToWork(\Laelaps\InvoiceBundle\Entity\Work $jobToWork)
{

    $this->jobToWork[] = $jobToWork;

    return $this;
}
/**
 * Remove jobToWork
 *
 * @param \Laelaps\InvoiceBundle\Entity\Work $jobToWork
 */
public function removeJobToWork(\Laelaps\InvoiceBundle\Entity\Work $jobToWork)
{
    $this->jobToWork->removeElement($jobToWork);
}

/**
 * Get jobToWork
 *
 * @return \Doctrine\Common\Collections\Collection
 */
public function getJobToWork()
{
    return $this->jobToWork;
}

../Entity/Work.php

/**
 * @ORM\ManyToOne(targetEntity="Job",inversedBy="jobToWork")
 * @ORM\JoinColumn(name="work_job", referencedColumnName="id")
 */
protected $workJob;

/**
 * Set workJob
 *
 * @param integer $workJob
 * @return Work
 */
public function setWorkJob($workJob)
{
    $workJob->addJobToWork($this);
    $this->workJob = $workJob;

    return $this;
}

/**
 * Get workJob
 *
 * @return integer
 */
public function getWorkJob()
{
    return $this->workJob;
}

I did a app/console doctrine:generate:crud

Which I know worked for my user to job relationship which was set up exactly the same way.

the error I get when I try to go to create a new "work" item is Object of class ..\Entity\Job could not be converted to string

Any help, as always greatly appreciated

UPDATE

As defined in the answer

public function __toString()
{
    return (string) $this->getJobName();
}

As I was using the FOSUserBundle and my user class extended the FOSUser class this meant that the __toString() function was defined in the FOSUser, hence there was no problem when creating assocations to user.

解决方案

You probably try to display your Job as a string (inside a select box or a list for example). Just add a __toString() method to your Job entity like this :

public function __toString()
{
    return $this->name; // For example
}

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