Symfony2可捕获的致命错误：DateTime类的对象无法转换为字符串

我有一个人（员工）搜索应该返回人名单，姓氏，出生日期和其他一些参数。对于日期参数（如出生日期），每个参数的搜索都可以正常工作。我的控制器代码是这样的：Symfony2可捕获的致命错误：DateTime类的对象无法转换为字符串

 $aWorker = new Worker(); 
    $searchWorkerForm = $this->createFormBuilder($aWorker) 
      ->add('omang', 'text', array('required' => false)) 
      ->add('workerName', 'text', array('required' => false)) 
      ->add('workerSurname', 'text', array('required' => false)) 
      ->add('birthDay', 'date', 
        array('required' => false, 'years' => range(1950, 2020))) 
      ->add('dateOfEmployment', 'date', array('required' => false)) 
      ->add('search', 'submit') 
      ->getForm(); 
    //Handle Request 
    if ($request->getMethod() == 'POST') 
     $searchWorkerForm->handleRequest($request); 

     $aWorkerList = $this->getDoctrine() 
      ->getRepository('OsdRetireBundle:Worker') 
      ->findByPersonDetails($aWorker->getOmang(), 
        $aWorker->getWorkerName(), 
        $aWorker->getWorkerSurname(), 
        $aWorker->getBirthDay(), 
        $aWorker->getDateOfEmployment()); 
//...

的findByPersonDetails功能是这样的：

public function findByPersonDetails($omang, $WorkerName, 
     $workerSurname, $birthDay, $dateOfEmployment){ 
    $qqq = $this->getEntityManager() 
      ->createQuery('SELECT w FROM OsdRetireBundle:Worker w 
       WHERE w.omang LIKE :omang 
       AND w.workerName LIKE :WorkerName 
       AND w.workerSurname LIKE :workerSurname 
       AND w.birthDay LIKE :birthDay 
       AND w.dateOfEmployment LIKE :dateOfEmployment') 
      ->setParameter('omang', '%'.$omang.'%') 
      ->setParameter('WorkerName', '%'.$WorkerName.'%') 
      ->setParameter('workerSurname', '%'.$workerSurname.'%') 
      ->setParameter('birthDay', '%'.$birthDay.'%') 
      ->setParameter('dateOfEmployment', '%'.$dateOfEmployment.'%') 
      ->getResult(); 
    return $qqq; 
}

的错误是：

Catchable Fatal Error: Object of class DateTime could not be converted to string in /var/www/Org/src/Osd/RetireBundle/Entity/WorkerRepository.php line 27 
500 Internal Server Error - ContextErrorException

我应该用我自己的函数将日期转换字符串？如果是的话我应该在哪里做到这一点，使其可重复使用？任何工作示例？我认为教义能够自动做到这一点。 Thak you in advenced。

来源

2013-10-02 Abel

[如何在Twig模板中渲染DateTime对象]的可能重复（http://stackoverflow.com/questions/8318914/how-to-render-a-datetime-object-in-a-twig-template ） – javidazac

DateTime对象没有实现__toString()方法，因此您必须将它明确地转换为字符串。

幸运的是，他们有一个format()方法，您可以利用来为你做这一点，例如：

->setParameter('birthDay', '%' . $birthDay->format('Y-m-d') . '%')

更多信息，请参见docs。

来源

2013-10-02 10:15:41 vascowhite

它的作品谢谢你们@Glavić和@vascowhite，我唯一需要做的就是确保鸟的日期不是空的。

public function findByPersonDetails($omang, $WorkerName, 
     $workerSurname, $birthDay, $dateOfEmployment){ 
    $createQuery = 'SELECT w FROM OsdRetireBundle:Worker w 
       WHERE w.omang LIKE :omang 
       AND w.workerName LIKE :WorkerName 
       AND w.workerSurname LIKE :workerSurname'; 
    if ($birthDay != '') 
     $createQuery .= ' AND w.birthDay LIKE :birthDay'; 

    if ($dateOfEmployment != '') 
     $createQuery .= 'AND w.dateOfEmployment LIKE :dateOfEmployment'; 

    $qqq = $this->getEntityManager() 
      ->createQuery($createQuery) 
      ->setParameter('omang', '%'.$omang.'%') 
      ->setParameter('WorkerName', '%'.$WorkerName.'%') 
      ->setParameter('workerSurname', '%'.$workerSurname.'%'); 
    if ($birthDay != '') $qqq = $qqq->setParameter('birthDay', '%'.$birthDay->format('Y-m-d').'%'); 
    if ($dateOfEmployment != '') $qqq = $qqq->setParameter('dateOfEmployment' 
      , '%'.$dateOfEmployment->format('Y-m-d').'%'); 
    $qqq = $qqq->getResult(); 
    return $qqq; 
}

我真的不知道我的代码看起来有多专业，但它适用于我。再次感谢您。

来源

2013-10-03 05:53:36 Abel

Symfony2可捕获的致命错误：DateTime类的对象无法转换为字符串

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