我在Mongo数据库中有以下文件：
db.totaldemands.insert({ "data" : "UKToChina", "demandPerCountry" :
{ "from" : "UK" , to: "China" ,
"demandPerItem" : [ { "item" : "apples" , "demand" : 200 },
{ "item" : "plums" , "demand" : 100 }
] } });
db.totaldemands.insert({ "data" : "UKToSingapore",
"demandPerCountry" : { "from" : "UK" , to: "Singapore" ,
"demandPerItem" : [ { "item" : "apples" , "demand" : 100 },
{ "item" : "plums" , "demand" : 50 }
] } });
我需要写一个查询来查找从英国出口到任何国家的苹果数量 .
我尝试了以下查询：
db.totaldemands.aggregate(
{ $match : { "demandPerCountry.from" : "UK" ,
"demandPerCountry.demandPerItem.item" : "apples" } },
{ $unwind : "$demandPerCountry.demandPerItem" },
{ $group : { "_id" : "$demandPerCountry.demandPerItem.item",
"total" : { $sum : "$demandPerCountry.demandPerItem.demand"
} } }
);
但它给了我苹果和李子的输出，如下所示：
{ "_id" : "apples", "total" : 300 }
{ "_id" : "plums", "total" : 150 }
但是，我的预期输出是：
{ "_id" : "apples", "total" : 300 }
So, How can I modify the above query to return only the count of apples exported from UK ? Also, is there any other better way to achieve the output without unwinding ?
1 回答
你可以添加另一个 $match 来获得苹果 .
由于您具有嵌入式文档结构和执行聚合，因此需要$ unwind . 备用选项可以是map和reduce . 但是，放松最适合这里 .
如果您正在考虑性能，则放松不应导致性能问题 .