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Laravel >$query->selectRaw'count*'->从'socket' ->leftJoin'container','socket.id','container.socket\u id' ->其中'socket.id',$socketId, “集装箱”_Laravel - Fatal编程技术网

Laravel >$query->selectRaw'count*'->从'socket' ->leftJoin'container','socket.id','container.socket\u id' ->其中'socket.id',$socketId, “集装箱”

Laravel >$query->selectRaw'count*'->从'socket' ->leftJoin'container','socket.id','container.socket\u id' ->其中'socket.id',$socketId, “集装箱”,laravel,Laravel,您可以尝试用toSql替换->get并转储结果,自己检查SQL。我使用这个查询来解决这个问题 $socjetsreport = DB::table('socket') ->select("socket.id", "socket.name", DB::raw("(SELECT COUNT(id) AS attchedcount FROM attached WHERE attached.socket_id=socket.id

您可以尝试用toSql替换->get并转储结果,自己检查SQL。

我使用这个查询来解决这个问题

$socjetsreport  = DB::table('socket')
    ->select("socket.id", "socket.name", 
    DB::raw("(SELECT COUNT(id) AS attchedcount FROM attached WHERE attached.socket_id=socket.id) as attchedcount"), 
    DB::raw("(SELECT COUNT(id) AS officecount FROM office WHERE office.socket_id=socket.id) as officecount"),
    DB::raw("(SELECT COUNT(id) AS containercount FROM container WHERE container.socket_id=socket.id) as containercount"),
    DB::raw("(SELECT COUNT(id) AS projectcount FROM project WHERE project.socket_id=socket.id) as projectcount"),
    DB::raw("(SELECT users.name AS employname FROM users WHERE users.id=socket.employee_id) as employname")
    )
    ->orderBy('socket.id', 'asc')
    ->get();

谢谢你的真棒