Php 如何使用不同的方法编写嵌套连接查询并将两列连接成一列在 laravel 查询生成器中

我最初的疑问是

    select distinct positions.pid, positions.eid, positions.name as position_name, elections.name as election_name, candidates.uname, firstname, surname 
from positions 
    inner join elections on positions.eid = elections.eid 
    inner join candidates on positions.pid = candidates.pid 
    inner join erole2 on candidates.uname = erole2.clogin 
    where elections.complaints_deadline <= CURRENT_TIMESTAMP;

我在laravel控制器中编写了上述查询，如下所示

$positions = DB::table('positions')
        ->join('elections', 'positions.eid', '=', 'elections.eid')
        ->join('candidates', 'positions.pid', '=', 'candidates.pid')
        ->join('erole2', 'candidates.uname', '=', 'erole2.clogin')
        ->select('positions.pid', 'positions.eid', 'positions.name as position_name', 'elections.name as election_name', 'candidates.uname', 'firstname', 'surname')
        ->where('elections.complaints_deadline', '<=', 'CURRENT_TIMESTAMP')
        ->distinct()->get();

But its not giving the same result. The original query returning 12 rows whereas the query I have written in my controller is returning an empty set. What i am doing wrong. How to write that query in laravel. I am using larvel 9.0个

第二，

在上面的代码中，我try 这样编写SELECT子句

 ->select('positions.pid', 'positions.eid', 'positions.name as position_name', 'elections.name as election_name', 'candidates.uname', '"CONCAT('firstname','surname') AS fullname"')

但是我收到错误消息，说Concat不是有效的列名.

你能帮我改正错误吗？