我正在使用PHP访问MongoDB集合，我在其中记录了游戏玩家:{username:"John",stats:{games_played:79,boosters_used:1,crystals:5}},{username:"Bill",stats:{games_played:0,boosters_used:0,crystals:20}},{username:"Jane",stats:{games_played:154,boosters_used:14,crystals:37}},{username:"Sarah",stats:{games_played:22,boosters_used
我正在尝试使用$sum计算每个州的总数。然后使用$lt仅显示小于100的状态。每笔总和:db.zipcode.aggregate([{$group:{_id:"$state",count_of_cities:{$sum:1}}}])尝试显示小于100的总和:db.zipcode.aggregate([{$group:{_id:"$state"},less_than:{$cond:{if:{$lt:[$sum:1,100]}}}}])样本收集:{"_id":"01001","city":"AGAWAM","loc":[-72.622739,42.070206],"pop":15338,"
我有一个聚合查询，它返回为给定位置提交的评论的总和/总数(不是平均星级)。评论评分为1-5星。这个特定的查询将这些评论分为两类，“内部”和“谷歌”。我有一个查询返回的结果几乎是我正在寻找的结果。但是，我需要为内部审查添加一个附加条件。我想确保内部评论“stars”值存在/不为null并且包含至少1的值。所以，我想添加类似这样的东西会起作用:{"stars":{$gte:1}}这是当前聚合查询:[{$match:{createdAt:{$gte:fromDate,$lte:toDate}}},{$lookup:{from:'branches',localField:'branch',fo
是否可以将$sum的结果添加到分组数组中？类似于:{"$group":{_id:{ProductId:"$ProductId",Day:"$Day"},Products:{$push:{clicks:{$sum:"$clicks"}}}}}我想将$sum的计算值存储到一个数组中。这可以分组完成吗？ 最佳答案 是的，您可以使用第二组运算符(operator)。db.collection.aggregate({$group:{_id:{ProductId:"$ProductId",Day:"$Day"},clicks:{$sum:"$c
在我的mongoDB管道的$facet阶段，因为一些记录引用相同的“customer.id”，我需要根据“customer”的unique出现次数进行计数。ID”。我试过这个:"totalCustomers":[{$group:{"$customer.id":{$sum:1}}}]...但最终出现错误:“字段名称‘$customer’不能是运算符(operator)名称”。我也试过这个:"totalCustomers":[{"$group":{"_id":"$customer.id","count":{"$sum":1}}}]...这不会出错，但会给出所有记录，而不是所有这些记录的计数
好的，所以我有一组看起来像这样的对象:{"address":{"building":"1007","coord":[-73.856077,40.848447],"street":"MorrisParkAve","zipcode":"10462"},"borough":"Bronx","cuisine":"Bakery","grades":[{"date":{"$date":1393804800000},"grade":"A","score":2},{"date":{"$date":1378857600000},"grade":"A","score":6},{"date":{"$date
我正在努力了解Mongo的整个聚合框架。我在这里有点新手。Ihavethefollowingdocuments:{"col1":"camera","fps":1,"lat":3},{"col1":"camera","fps":3,"lat":2}{"col1":"recorders","fps":9,"lat":7}{"col1":"cell","fps":8,"lat":1}{"col1":"cell","fps":4,"lat":3}如何设置返回结果:{"col1":"camera","fps":4,"lat":5},{"col1":"recorders","fps":9,"lat
我有以下收藏:error_reports[{"_id":{"$oid":"5184de1261"},"date":"29/04/2013","errors":[{"_id":"10","failures":2,"alerts":1,},{"_id":"11","failures":7,"alerts":4,}]},{"_id":{"$oid":"5184de1262"},"date":"30/04/2013","errors":[{"_id":"15","failures":3,"alerts":2,},{"_id":"16","failures":9,"alerts":1,}]}]是
这是我的收藏:{"_id":10926400,"votes":131,"author":"Jesse","comments":[{"id":1,"votes":31,"author":"Mirek"},{"id":2,"votes":13,"author":"Leszke"}]},{"_id":10926401,"votes":75,"author":"Mirek","comments":[{"id":1,"votes":17,"author":"Jesse"},{"id":2,"votes":29,"author":"Mirek"}]}我想要每个author的$sumvotes和co
我有一个像这样的聚合函数:varmatch={};match["products.totalprice"]={$exists:true};varproject={};project["_id"]=0project["products.totalprice"]=1;project["line"]="$products.closedate";ThisCollection.aggregate([{$match:match},{$project:project},{$group:{_id:"$line",total:{$sum:{$ifNull:["$products.totalprice",
