PHP錯誤調用一個成員函數格式（）上陣列

Question

我有一個日期時間字段

/** 
* @var \DateTime 
* 
* @ORM\Column(name="date", type="datetime", nullable=true) 
*/ 
private $datetime; 

/** 
* Set date 
* 
* @param \DateTime $datetime 
* 
*/ 
public function setDate($datetime) 
{ 
    $this->datetime = $datetime; 

    return $this->datetime ?? new \DateTime(); 
} 

/** 
* Get date 
* 
* @return \DateTime 
*/ 
public function getDate(): \DateTime 
{ 
    return $this->datetime ?? new \DateTime(); 
} 

我此錯誤下面：

呼叫一個成員函數格式（）上陣列

任何人都知道我爲什麼得到這個？

編輯：

下面是我用它來生成表單的日期時間值讀取，然後收集從表單中的數據，並在表中創建一個新的實體代碼：

$trainingform = new Training(); 

    $form = $this->createFormBuilder($trainingform) 
     ->add('Leader', TextType::class) 
     ->add('Date', DateTimeType::class, ['label' => 'Date and Time']) 
     ->add('topics', TextType::class, ['label' => 'Topics Being Covered']) 
     ->getForm(); 

    if ($form->handleRequest($request)->isValid()) { 
     $trainingform->setLeader($request->request->get('form')['Leader']); 
     $trainingform->setDate($request->request->get('form')['Date']); 
     $trainingform->setTopics($request->request->get('form')['topics']); 
     $em->persist($trainingform); 
     $em->flush(); 
    } 

Answer 1

爲了正確處理您的數據，您應該使用symfony表單的getData-方法，它會將您的請求轉換爲您之前在表單中指定的狀態。在你的情況應該是這樣的：

/** 
* Set date 
* 
* @param \DateTime $datetime 
*/ 
public function setDate(\DateTime $datetime) //typhint added 
{ 
    $this->datetime = $datetime; 
} 

`

二傳手不應該返回任何東西：

$trainingForm = new Training(); 

    $form = $this->createFormBuilder($trainingForm) 
     ->add('Leader', TextType::class) // 'Leader' should be named as your property 
     ->add('datetime', DateTimeType::class, ['label' => 'Date and Time']) //changed your Date to datetime as your property is 
     ->add('topics', TextType::class, ['label' => 'Topics Being Covered']) //should be named as your property as well 
     ->add('submit', SubmitType::class, ['label' => 'Submit form']) //should be there 
     ->getForm(); 

    $form->handleRequest($request) 

    if (form->isSubmitted() && form->isValid()) { 
     $trainingForm= $form->getData(); 
     $em->persist($trainingForm); 
     $em->flush(); 
    } 

爲了避免這樣的不一致，你可以在你的setter方法指定所需的類型順便說一句。然後，你會在你可以控制的步驟上得到你的錯誤，你會知道你應該在哪裏挖。