Question

我在Laravel 5.4上遇到问题，当我尝试只使用一个连接时，它可以正常工作并返回正确的数据，但是他们添加了另一个不起作用的连接。

$data = Player::select(DB::raw('CONCAT(familyName,", ",firstName) AS fullName'))
    ->where('firstname', 'like', '%'.$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)

Answer 1

如果你要加入表格，那么你应该给表别名。比如team as t，player为p，然后是列名，如p.playername

Laravel加入多个表不起作用

1 个答案: