仅显示当前用户的关系

Question

我有2个实体，用户和课程，每个学生都可以有很多课程，每个课程都可以有很多学生。无论如何，我正在尝试做到这一点，以便当用户连接到他的帐户时，它向他展示了与他相关的所有课程。我知道我必须从控制器中做到这一点，但是我只能得到他与之相关的人。我的实体

<?php
/**
 * @ORMEntity
 */
class User implements UserInterface
{
    /**
     * @ORMId
     * @ORMColumn(type="integer")
     * @ORMGeneratedValue(strategy="AUTO")
     */
    private $id;
    public function getId()
    {
        return $this->id;
    }

    /**
     * @ORMManyToMany(targetEntity="AppEntityCourses", mappedBy="Users")
     */
    private $courses;
    public function __construct()
    {
        $this->courses = new ArrayCollection();
    }

    /**
     * @return Collection|course[]
     */
    public function getCourses(): Collection
    {
        return $this->courses;
    }
    public function addCourse(Course $course): self
    {
        if (!$this->courses->contains($course)) {
            $this->courses[] = $course;
            $course->addUser($this);
            return $this;
        }
        return $this;
    }
    public function removeCourse(Course $course): self
    {
        if ($this->courses->contains($course)) {
            $this->courses->removeElement($course);
            $course->removeUser($this);
        }
        return $this;
    }
}

<?php
/**
 * @ORMEntity(repositoryClass="AppRepositoryCourseRepository")
 */
class Course
{
    /**
     * @ORMId()
     * @ORMGeneratedValue()
     * @ORMColumn(type="integer")
     */
    private $id;
    /**
     * @ORMColumn(type="string", length=55)
     */
    private $name;
    /**
     * @ORMManyToMany(targetEntity="AppEntityuser", inversedBy="courses")
     */
    private $users;

    public function __construct()
    {
        $this->users = new ArrayCollection();
    }
    public function getId(): ?int
    {
        return $this->id;
    }
    public function getname(): ?string
    {
        return $this->name;
    }
    public function setname(string $name): self
    {
        $this->name = $name;
        return $this;
    }

    /**
     * @return Collection|user[]
     */
    public function getUser(): Collection
    {
        return $this->user;
    }
    public function addUser(user $user): self
    {
        if (!$this->users->contains($user)) {
            $this->users[] = $user;
        }
        return $this;
    }
    public function removeUser(user $user): self
    {
        if ($this->users->contains($user)) {
            $this->users->removeElement($user);
        }
        return $this;
    }

}

Answer 1

好吧，因为您显然有双向关系(反向和mappedby(，您无需做任何特别的事情。学说已经做到了。只需在控制器中抓住用户：

// Once you have a protected controller route,
// you can use $this->getUser() from inside the
// controller action to access the current authenticated user.
$user = $this->getUser();

和您想要通过他们的相关课程的迭代：

foreach ($user->getCourses() as $course) {
}

因为getCourses还返回了Collection实例，因此您还可以使用->map和->filter函数，该功能与收集类一起运送...