Question

我有以下用户架构

{
        "__v" : 26,
        "_id" : ObjectId("52b47058fe5e3493a2cf8365"),
        "live_sockets" : [
                "CEGok0rSz2UtBIHX9DlV",
                "s6M45OA0MDBs4aGL9DlW",
                "2XiszSuyiPAGr-Ga9ZCN",
                "lIFNEeUgXRB6tgvP9ZCO",
                "JPtzQOD_52maf6VS9gtb",
                "kDL06aXiI8WWlig19gtc",
                "75Bt5p6WqgmRcyer9xSm",
                "Ge_sKLen32Q91wLW9xSn",
                "EpHt3qju34GTZevU_Bsd",
                "n0hq0EQAjJOptxdy_qEB",
        ],
        "name" : {
                "first" : "Test",
                "last" : "User"
        },
        "notifications_count" : 0,
        "role" : 1
}

如何基于特定的实时套接字查询用户，因为socket_ids是唯一的。如果我有一个名为current_socket_id的变量，我怎样才能找到套接字所属的用户..？

Answer 1

db.collection.find({"live_sockets" : {$in : [socketId]}},{"name" : 1,"_id":0})

将给出套接字ID的名称

如何根据mongoose中字段数组的值查询集合

1 个答案: