使用 laravel 查询生成器连接两个表无法获得结果

Question

我有两个表'用户'和'频道'表：用户id  name        channel1   user1       1,2,32   user2       2,33   user3       2表：渠道id   channel_name1    IT2    CS3    EC我需要结果name     channel_nameuser1    IT,CS,ECuser2    CS,ECuser3    CS使用 laravel 查询生成器如何编写查询？我在下面尝试过，但我将channel_name 设置为NULL。试试 1$UserChannelList = Users::select('users.name as username', DB::raw("(GROUP_CONCAT(channels.channel_name SEPARATOR ',')) as 'channel_name'"))        ->leftjoin('channels', function ($join) {          $join->whereRaw("FIND_IN_SET('channels.id', 'users.channel')");        })        ->groupBy('users.name')        ->orderBy('users.name', 'ASC')        ->get();试试 2$UserChannelList = Users::select('users.name as username', DB::raw("(GROUP_CONCAT(channel.channel_name SEPARATOR ',')) as 'channel_name'"))        ->leftjoin('channel', function ($join) {          $join->on(DB::raw("CONCAT(',', 'users.channel', ',')"), 'like', DB::raw("CONCAT(',','channel.id',',')"));        })        ->groupBy('users.name')        ->orderBy('users.name', 'ASC')        ->get();

Answer 1

试试这个查询。

\DB::table("users")
        ->select("users.*",\DB::raw("GROUP_CONCAT(channels.channel_name) as channel_name"))
        ->leftjoin("channels",\DB::raw("FIND_IN_SET(channels.id,users.channel)"),">",\DB::raw("'0'"))
        ->get();

请检查我的回答

Answer 2

这是您的解决方案，请检查

    $UserChannelList = DB::table('users')
                        ->select('users.name', DB::raw("(GROUP_CONCAT(channels.channel_name)) as 'channel_name'"))
                        ->rightJoin('channels', function($join){                            $join->whereRaw('FIND_IN_SET(channels.id, users.channel)');
                        })
                        ->groupBy('users.name')
                        ->orderBy('users.name', 'ASC')
                        ->get();