Question

我有这个代码

> db.attributes.aggregate([{ $match: {"name.en-UK": "Size"} },{ $unwind : "$values" },{ $project: { "code" : "$values.code", "name" : "$values.name", "_id": "$values._id"} },{ $match: {"name.en-UK": "S"}}])
{
    "result" : [
        {
            "_id" : ObjectId("53b2c644d901f244965f74e3"),
            "code" : 0,
            "name" : {
                "en-UK" : "S"
            }
        },
        {
            "_id" : ObjectId("53b2c644d901f244965f74e4"),
            "code" : 1,
            "name" : {
                "en-UK" : "S"
            }
        }
    ],
    "ok" : 1
}

如何更改它以便它只返回＆＃34;代码＆＃34; == 0 with＆＃34; name＆＃34; ==＆＃34; en-UK＆＃34; ：＆＃34; S＆＃34;

我试过这个：

> db.attributes.aggregate([{ '$match': {"name.en-UK": "Size"} },
... { '$unwind' : "$values" },
... { '$project': { "code" : "$values.code", "valueId": "$values._id"} },
... { '$match': {
... $and: [
... {"name.en-UK": "S"},
... {"code": 0}
... ]
... }
... }])
{ "result" : [ ], "ok" : 1 }

但是没有用！

和建议非常感谢

Answer 1

似乎我需要向select添加内容，因此{ $project: { "code" : "$values.code", "name" : "$values.name", "_id": "$values._id"} }必须包含名称

> db.attributes.aggregate([{ '$match': {"name.en-UK": "Size"} }, { '$unwind' : "$values" }, { $project: { "code" : "$values.code", "name" : "$values.name", "_id": "$values._id"} }, { '$match': { $and: [  {"name.en-UK": "S"},  {"code": 0} ] } }])
{
    "result" : [
        {
            "_id" : ObjectId("53b2c644d901f244965f74e3"),
            "code" : 0,
            "name" : {
                "en-UK" : "S"
            }
        }
    ],
    "ok" : 1
}

mongodb和聚合多个项目

1 个答案: