如何將表列值與Auth :: user-> id在Laravel中進行比較

Question

我的問題是這個我正在使用下面的代碼來訪問不同角色的系統用戶。

public function show($id) 
{ 
    if (Permission::where('status', 1)->where('project_id', $id)->exists()) { 
    // if((Permission::where('status', '=', '1')->first()) && (Permission::where('project_id','=',$id)->first())){ 
     $project = Project::find($id); 
     $tasks = $this->getTasks($id); 
     $files = $this->getFiles($id); 
     $comments = $this->getComments($id); 
     $collaborators = $this->getCollaborators($id); 
     $permissions = $this->getPermissions($id); 
returnview('collaborators.show')->withProject($project)->withTasks($tasks)->withFiles($files)->withComments($comments)->withCollaborators($collaborators); 
     } 
    else if 
     //return('hi'); 
     (Permission::where('status', 2)->where('project_id', $id)->exists()) { 
      $project = Project::find($id); 
      $tasks = $this->getTasks($id); 
      $files = $this->getFiles($id); 
      $comments = $this->getComments($id); 
      $collaborators = $this->getCollaborators($id); 
      $permissions = $this->getPermissions($id); 
      return view('collaborators.manager')->withProject($project)->withTasks($tasks)->withFiles($files)->withComments($comments)->withCollaborators($collaborators); 
     } 

在我的權限表有collaborator_id，這是相同的用戶表的用戶ID。我需要使用登錄用戶Auth::user->id驗證（比較）collaborator_id。在以下腳本中。

if (Permission::where('status', 1)->where('project_id', $id)->exists()) 

怎麼能行呢？

Answer 1

where子句只是用另一種

Permission::where('status', 1)->where('project_id', $id)->where('collaborator_id', Auth::User()->id) 

建議：您可以通過避免if和else相同/重複代碼減少你的代碼。