选择不在另一个表laravel 5.5中的所有列

DB::table("users")->select('*')->whereNotIn('user_name',function($query) {

   $query->select('user_name')->from('buy_courses');

})->get();

只是 实际上是拉腊维尔的内部连接,所以实际上你也可以尝试:

DB::table('users')
            ->join('buy_courses', 'users.user_name', '=', 'buy_courses.user_name')
            ->get();

试着使用雄辩的:

$courseUserNames = BuyCourses::pluck('user_name')->all();
$users = User::whereNotIn('user_name', $courseUserNames)->select(...)->get();

或者,如果您更喜欢使用DB查询:

$courseUserNames = DB::table('buy_courses')->pluck('user_name')->all();
$users = DB::table('users')->whereNotIn('user_name', $courseUserNames)->select(...)->get();

= 具有 !=

$users = DB::table('users')
  ->join(
    'buy_courses', 
    function ($join)
    {$join->on('users.user_name', '!=', 'buy_courses.user_name');}
  )
  ->get();

例子:

mysqli_query($con, "SELECT * FROM users WHERE user_name NOT IN (SELECT user_name FROM buy_courses)");