Question

对于NoSQL的兴高采烈我还是有点陌生，我从一个以MongoDB作为数据库的项目开始。

目标：打印至少有3个等级的住宿清单，大多数等级（半个或更多）都高于6。内部住宿。

问题：我有很多评论。我想将这些对象合并成一个等级数组，并从每个对象中排除审阅者和日期。

整理完毕：然后，我将$ gt大于5的对象（所以6个或更多）推到isGtArray，并将wersa推到isLtArray。然后，我可以算出是否确实有一半以上的分数高于6，并在isMostOfTheGradesMoreThan_6中进行投影，并使用$ match将它们打印出来。到现在为止：解决了住宿等级至少为3 的第一部分。非常感谢您的帮助。

现在输入代码：

db.lodging.aggregate([
{$unwind: "$lodging.reviews"},
{$group:{
    _id:"$guid",
    lodging: { $first: "$lodging"},
    reviews: { $push: "$lodging.reviews"}}
},
{$project: { 
    _id: 1,
    lodging: 1,
    reviews: 1,
    reviewsCount: { $size:"$reviews" }}},
{ $match : { reviewsCount: { $gt: 2 } } },
])

Json数据

{ 
    "_id" : "5bf31f8c20b3acacc2194faa", 
    "registered" : "2016-04-22T08:54:50 -02:00", 
    "host" : {
        "name" : "Schultz", 
        "surname" : "French",  
        "contact" : {"email" : "schultzfrench@orbean.com"}, }, 
    "lodging" : { 
        "reviews" : [
             {
                "reviewer" : "Mariana Ayers", 
                "date" : "2014-04-13T08:30:02 -02:00", 
                "cleanliness" : NumberInt(9), 
                "location" : NumberInt(6), 
                "food" : NumberInt(4)}, 
             {
                "reviewer" : "Cole Simmons", 
                "date" : "2015-05-07T06:00:09 -02:00", 
                "cleanliness" : NumberInt(1), 
                "location" : NumberInt(3), 
                "food" : NumberInt(2)}]}
}

解决方案

{ 
    "_id" : "5bf31f8c20b3acacc2194faa", 
    "registered" : "2016-04-22T08:54:50 -02:00", 
    "host" : {
        "name" : "Schultz", 
        "surname" : "French",  
        "contact" : {"email" : "schultzfrench@orbean.com"},}, 
    "lodging" : { 
        "reviews" : [
            {
                "reviewer" : "Mariana Ayers", 
                "date" : "2014-04-13T08:30:02 -02:00", 
                "cleanliness" : NumberInt(9), 
                "location" : NumberInt(6), 
                "food" : NumberInt(4)}, 
            {
                "reviewer" : "Cole Simmons", 
                "date" : "2015-05-07T06:00:09 -02:00", 
                "cleanliness" : NumberInt(1), 
                "location" : NumberInt(3), 
                "food" : NumberInt(2)}]
        "reviews_grades" : [
                {"cleanliness" : NumberInt(9)}, 
                {"location" : NumberInt(6)}, 
                {"food" : NumberInt(4)},
                {"cleanliness" : NumberInt(1)}, 
                {"location" : NumberInt(3)}, 
                {"food" : NumberInt(2)}}]}
}

MongoDB，合并数组中的对象

0 个答案: