mongoose踩坑记

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Question1:返回集合【数组有至少两个以上相同的值】

集合数据如下：

{     "_id" : ObjectId("5a28e0b6eeb0b8783a382bd0"),     "name" : "a",     "activeClass" : [        {            "status" : 0.0        },         {            "status" : 8.0        },         {            "status" : 0.0        }    ]}{     "_id" : ObjectId("5a28e0d2eeb0b8783a382bd1"),     "name" : "b",     "activeClass" : [        {            "status" : 8.0        },         {            "status" : 8.0        },         {            "status" : 0.0        }    ]}

查询条件：

db.collection.aggregate(    {$unwind: "$activeClass" },    {$match: {"activeClass.status":8}},    {$group: {"_id": "$_id", "count": {"$sum": 1}}},    {$match: {count :{$gte :2}}});

返回结果：

{     "_id" : ObjectId("5a28e0d2eeb0b8783a382bd1"),     "count" : 2.0}

解析：
unwind 是根据 $activeClass中的元素分解成单个。

查询：db.collection.aggregate(    {$unwind: "$activeClass" });返回结果：{     "_id" : ObjectId("5a28e0b6eeb0b8783a382bd0"),     "name" : "a",     "activeClass" : {        "status" : 0.0    }}{     "_id" : ObjectId("5a28e0b6eeb0b8783a382bd0"),     "name" : "a",     "activeClass" : {        "status" : 8.0    }}{     "_id" : ObjectId("5a28e0b6eeb0b8783a382bd0"),     "name" : "a",     "activeClass" : {        "status" : 0.0    }}{     "_id" : ObjectId("5a28e0d2eeb0b8783a382bd1"),     "name" : "b",     "activeClass" : {        "status" : 8.0    }}{     "_id" : ObjectId("5a28e0d2eeb0b8783a382bd1"),     "name" : "b",     "activeClass" : {        "status" : 8.0    }}{     "_id" : ObjectId("5a28e0d2eeb0b8783a382bd1"),     "name" : "b",     "activeClass" : {        "status" : 0.0    }}

Question2:查询条件如何表达数组指定元素

集合数据：

{     "_id" : ObjectId("5a28e0b6eeb0b8783a382bd0"),     "name" : "a",     "activeClass" : [        {            "status" : 0.0        },         {            "status" : 8.0        },         {            "status" : 0.0        }    ]}

查询：返回activeClass数组第一个元素值为1的集合；

db.collection.find({activeClass.0.status: 1});

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