如何从表中获取数据并进行比较并在laravel中显示结果？

Question

我想从表中获取数据。我有两张表，一个测试，另一个是用户，我想将技能与技能进行比较，并显示与技能和技能相匹配的工作。我该怎么办？

public function post( Request $request)
{
    Test::create([
       'jobtitle'=>$request['jobtitle'],
       'company'=>$request['cname'],
       'jdesc'=>$request['jdesc'],
       'skillneed'=>$request['Skills'],
       'duration'=>$request['duration'],
       'qualification'=>$request['qualification'],
       'experiance'=>$request['experiance'],
      'location'=>$request['location'],
    ]);
}



    protected function create(array $data)
    {
       return User::create([
            'name' => $data['name'],
            'email' => $data['email'],
            'password' => Hash::make($data['password']),
            'mobile' => $data['mobile'],
            'age' => $data['age'],
            'location' => $data['location'],
            'resume' => $data['resume'],
            'skills' => $data['skills'],
            'course' => $data['course'],
            'college' => $data['college'],
            'role_id'=>$data['role_id'],

        ]);
    }

Answer 1

您好，您可以使用collection helper contains

$testCollection->contains('skillneed', $user->skills)返回布尔值

找到所有测试

$filtered = $testCollection->where('skillneed', $user->skills);返回已过滤的集合

或与DB

$skills = Test::where('skillneed', $user->skills)->get()

Answer 2

这解决了我的问题

$skill = Auth::user()->skills;
$res = DB::table('tests')->where('skillneed',$skill)->get();
return view('jobs.jobmatching',compact('res'));