Php Laravel联接多个表不工作

我在Laravel5.4上遇到了一个问题，当我尝试只使用一个连接时，它工作正常并返回正确的数据，但是他们的“添加另一个连接”不起作用

$data = Player::select(DB::raw('CONCAT(familyName,", ",firstName) AS fullName'))
    ->where('firstname', 'like', '%'.$search.'%')
    ->orWhere('familyName', 'like', '%'.$search.'%')
    ->orderBy('familyName', 'asc')
    ->join('teams', 'players.primaryClubId', '=', 'teams.clubId')
    ->join('person_competition_statistics', 'players.personId', '=', 'person_competition_statistics.personId')
    ->addSelect(['players.*', 'teams.teamName', 'teams.teamNickname', 'teams.teamCode'])
    ->get()
    ->unique() //remove duplicates
    ->groupBy(function($item, $key) { //group familyName that starts in same letter
        return substr($item['familyName'], 0, 1);
    })
    ->map(function ($subCollection) {
        return $subCollection->chunk(4); // put your group size
    });

return $data;

返回的错误：

QueryException in Connection.php line 647:
SQLSTATE[23000]: Integrity constraint violation: 1052 Column 'familyName' in field list is ambiguous (SQL: select CONCAT(familyName,", ",firstName) AS fullName, `players`.*, `teams`.`teamName`, `teams`.`teamNickname`, `teams`.`teamCode` from `players` inner join `teams` on `players`.`primaryClubId` = `teams`.`clubId` inner join `person_competition_statistics` on `players`.`personId` = `person_competition_statistics`.`personId` where `firstname` like %% or `familyName` like %% order by `familyName` asc)

---

若要加入表，则应提供表别名。就像团队是t，球员是p，然后列名是p.playername

在你要加入的三个表中，是否有多个表中的列

familyName

？如果是这样，则需要在查询中使用表名的任何地方都使用表名作为前缀。像

DB:：raw（'CONCAT（players.familyName，，，，，firstName）作为全名）

谢谢@ayip的帮助，现在就到了working@PenAndPapers若要加入表，则应提供表别名。比如团队是t，球员是p，然后列名是p@NikhilRadadiya@PenAndPapers工作？