mongo in（）子句按大多数匹配排序

Question

问题显示如下：

我的疑问：

db.goods.find({tags:{$in:["white","black","gray"]}}).pretty();

并返回数据：

{
    "id":1,
    "tags": [
        "black",
        "blue"
    ]
}
{
    "id":2,
    "tags": [
        "white",
        "gray"
    ]
}
{
    "id":3,
    "tags": [
        "gray",
        "black",
        "white"
    ]
}

现在我希望通过标记的显示来排序查询，即_id：3是标记字段的最匹配记录，后跟_id：2。

有人可以帮我解决这个问题吗？

Answer 1

您可以使用以下聚合查询来获得所需的结果。

db.goods.aggregate([
    {$match: {tags: {$in: ["white","gray","black"]}}}, 
    {$project: {"tags":1, "tagsCopy":"$tags"}},
    {$unwind: "$tagsCopy"}, 
    {$match: {tagsCopy: {$in: ["white","gray","black"]}}},
    {$group: {
        _id:"$_id", 
        counter:{$sum:1},
        tags:{"$first":"$tags"}
    }},
    {$sort:{counter:-1}},   
    {$project: {"tags":1}}
]);

Answer 2

可以使用聚合

解决

db.goods.aggregate([
    {$match: {tags: {$in: ["white","gray","black"]}}}, 
    {$unwind: "$tags"}, 
    {$match: {tags: {$in: ["white","gray","black"]}}},
    {$group: {
        _id:"$_id", 
        matches:{$sum:1},
        tags: {$push:"$tags"}
    }}, 
    {$sort:{matches:-1}}
]);