如何在Laravel中用口才来获取用户的myProjects任务

Question

我是Laravel和Eloquent的新手。在我的应用程序中，我想检索当前用户和其他项目成员的任务，并将其显示在视图中。这是我建立的数据库关系：

users（id）

projects（ID，用户ID）

user_project（ID，用户ID，项目ID，角色）

tasks（ID，用户ID，项目ID）

（（我已经定义了迁移中所需的外键）

为了解释这种关系，每个项目都有一个用户（创建该项目的用户）。但是在user_project表中，也可以将用户分配给其他项目，并在其中定义他们的角色。例如：有两个用户：id = 1id = 2

和三个项目：id = 1，user_id = 1id = 2，user_id = 2id = 3，user_id = 1

user_project关系：id = 1，user_id = 1，project_id = 1，role = adminid = 2，user_id = 2，project_id = 1，role = employeeid = 3，user_id = 2，project_id = 2，role = adminid = 4，user_id = 1，project_id = 3，role = admin

还有四个任务：id = 1，user_id = 1，project_id = 1id = 2，user_id = 2，project_id = 1id = 3，user_id = 1，project_id = 2id = 4，user_id = 1，project_id = 3

我希望用户id=2能够看到project_id=1的任务，因为他已被邀请作为员工加入该项目，而用户project_id=2的任务也已被创建，因为他已经创建了该项目。当然，由于用户不是成员，因此用户不应看到project_id = 3的任务。那么，最简​​洁的方法是什么？

这里是我定义的模型：

class User extends Authenticatable
{
    public function projects(){
        return $this->hasMany(Project::class);
    }
    public function tasks(){
        return $this->hasMany(Task::class);
    }
    public function joinedProjects()
    {
        return $this->hasMany(ProjectUser::class);
    }
}

class Project extends Model
{
    public function tasks(){
        return $this->hasMany(Task::class);
    }
    public function users(){
        return $this->hasMany(User::class);
    }
}

class ProjectUser extends Model
{
}

class Task extends Model
{
    public function projects(){
        return $this->belongsTo(Project::class);
    }
    public function users(){
        return $this->belongsTo(User::class);
    }
}

这是我尝试检索项目成员任务的方式（我想要实现这一目标的理想方法是：$tasks = $user->joinedProjects->tasks，但我不知道如何做到这一点，所以这就是我当前正在尝试完成此操作）：

class TasksController extends Controller
{
    public function index()
    {
        $user = Auth()->user();
        //I guess there are better ways to retrieve projects, but:
        $projects = app('App\Http\Controllers\HomeController')->allProjects($user);
        foreach($projects as $project){
            $tasks[] = Task::where('project_id', $project->id);
        }
        return view('tasks.index', compact(['tasks','user']));
        //gives error: Property [id] does not exist on the Eloquent builder instance
        //when I'm trying to get $task->id in a foreach loop.
    }
}

这是家庭控制器（对于其他一些类功能，我需要HomeController中的allProjects()函数：]

class HomeController extends Controller
{
    function allProjects($user){
        $projects = $user->projects;
        $otherProjects = \App\ProjectUser::where('user_id',$user->id)->get();
        foreach ($otherProjects as $project){
            $projects[] = \App\Project::find($project->project_id);
        }
        return $projects;
    }
}

Answer 1

首先，我认为您应该将joinedProjects设置为many-to-many relationship，这样访问它会更简单。

// in User model
public function joinedProjects()
{
    // i'm assuming you want to always have access to the role property
    return $this->belongsToMany(Project::class, 'user_project')->withPivot('role');
}

// in Project model
public function memberUsers()
{
    return $this->belongsToMany(User::class, 'user_project')->withPivot('role');
}

通过这种关系，您应该可以调用$user->joinedProjects以获得用户已加入的项目列表。

当然，您可以调用joinedProjects关系并像生成的for循环那样循环遍历生成的项目。或者，您也可以使用集合类的pluck方法。

class TasksController extends Controller
{
    public function index()
    {
        // here I put load to eager load the project and task
        $user = Auth()->user()->load('joinedProjects.tasks');

        // OPTION 1: you can loop to get the tasks
        $tasks = collect();
        foreach($user->joinedProjects as $project){
            $tasks = $tasks->merge($project->tasks);
        }

        // OPTION 2: or use pluck (must be eager loaded to work)
        $tasks = $user->joinedProjects->pluck('tasks');

        // $tasks should be unique here, but if it is not you can call the unique method of collection
        $tasks = $tasks->unique('id');

        return view('tasks.index', compact(['tasks','user']));
    }
}

您共享的HomeController也可以通过新的关系简化

class HomeController extends Controller
{
    function allProjects($user){
        // here i'm assuming user automatically joins a project when they create it
        $projects = $user->joinedProjects;

        return $projects;
    }
}

这里是我在代码和unique中使用的lazy eager loading方法的一些附加参考信息>