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我正在使用php7的类型提示。所以我下面的代码类型错误:返回值必须是实例,空返回
class Uni
{
/**
* @var Student
*
* @ORM\ManyToOne(targetEntity="AppBundle\Entity\Student")
* @ORM\JoinColumn(nullable=false)
*/
private $student;
function _construct(){
$this->student= new Student();
}
/**
* Set student
*
* @param \AppBundle\Entity\Student $student
*
* @return Uni
*/
public function setStudent (Student $student): Uni
{
$this->student= $student;
return $this;
}
/**
* Get student
*
* @return \AppBundle\Entity\Student
*/
public function getStudent(): Student
{
return $this->student;
}
}
现在,当我尝试加载统一新形式,我得到这个错误
Type error: Return value of AppBundle\Entity\Uni::getStudent() must be an instance of AppBundle\Entity\Student, null returned
我怎样才能摆脱这种错误的?我发现了一个可空的解决方案,它需要php 7.1。但现在我必须坚持使用PHP 7.0。那我该如何解决这个问题?