在我的Laravel5.7/雄辩/MySQL5.5应用程序中,我需要在聊天列表中获得任何聊天的消息的值,并为给定的用户($user_id)多获得一列聊天消息:
我试过这样做:
$chats = Chat
::select( \DB::raw( ' chats.*, count(cm.id) as messages_count, count(cmu.id) as user_messages_count' ) )
->orderBy('name', 'asc')
->groupBy('chats.id')
->leftJoin( \DB::raw('chat_messages as cm'), \DB::raw('cm.chat_id'), '=', \DB::raw('chats.id') )
->leftJoin( \DB::raw('chat_messages as cmu'), \DB::raw(' cmu.chat_id = chats.id and cmu.chat_id = ' .$user_id ) )
->get(); 但错误:
Column not found: 1054 Unknown column '' in 'on clause'.看来leftJoin表达式不能与空值参数一起使用。还是以其他方式?
$chats = Chat
::select( \DB::raw( ' chats.*, count(cm.id) as messages_count, count(cmu.id) as user_messages_count' ) )
->orderBy('name', 'asc')
->groupBy('chats.id')
->leftJoin( \DB::raw('chat_messages as cm'), \DB::raw('cm.chat_id'), '=', \DB::raw('chats.id') )
->leftJoin(\DB::raw('chat_messages as cmu'), \DB::raw('cmu.chat_id'), '=', \DB::raw('chats.id'))
->where( 'cmu.user_id', $user_id)
->get(); // `chat_messages` ORDER BY `chat_id` A这是可行的,但就条件而言却不正确。
>where( 'cmu.user_id', $user_id)跳过左联接,跳过带有user_messages_count =0的行,但我需要它们。
哪条路是对的?
谢谢!
发布于 2018-12-29 12:12:47
您不需要多次向左加入chat_messages。在这种情况下,WHERE子句是无用的。LEFTJOIN处理连接表中的列。
$chats= Chat ::select( \DB::raw( ' chats.*, count(cm.id) as
messages_count, sum(if(cm.user_id='.$user_id.',1,0)) as user_messages_count' ) )
->orderBy('name', 'asc')
->groupBy('chats.id')
->leftJoin( \DB::raw('chat_messages as cm'), \DB::raw('cm.chat_id'), '=', \DB::raw('chats.id') )
->get(); // `chat_messages` ORDER BY `chat_id` Ahttps://stackoverflow.com/questions/53969187
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