mongoDB:如何反转 $unwind

Question

考虑这组测试结果:

[{
    _id: ObjectId(...),
    name: "Test1",
    acts: [
    {
        name: "act1", 
        tests: [
            {name: "test1", result: true}, 
            {name: "test2", result: true}]
    }]
},
{
    _id: ObjectId(...),
    name: "Test2",
    acts: [
    {
        name: "act1", 
        tests: [
            {name: "test1", result: true}, 
            {name: "test2", result: false}]
    }, 
    {
        name: "act2", 
        tests: [
            {name: "test3", result: true}]
    }]
}]

我正在尝试使用聚合来创建一个包含所有测试结果总和的计算字段，我想要这样的东西:

[{
    _id: ObjectId(...),
    name: "Test1",
    result: true, //new aggregated value
    acts: [
    {
        name: "act1", 
        result: true, //new aggregated value
        tests: [
            {name: "test1", result: true}, 
            {name: "test2", result: true}]
    }]
},
{
    _id: ObjectId(...),
    name: "Test2",
    result: false, //new aggregated value
    acts: [
    {
        name: "act1", 
        result: false, //new aggregated value
        tests: [
            {name: "test1", result: true}, 
            {name: "test2", result: false}]
    }, 
    {
        name: "act2", 
        result: true, //new aggregated value
        tests: [
            {name: "test3", result: true}]
    }]
}]

我尝试过使用 aggregate 和 $unwind、$project 和 $group:

aggregate([
  {$unwind: "$acts"},
  {$unwind: "$acts.tests"},
  {$project: {name: 1, acts: 1, failed: {$cond: {if: {$eq: ["$acts.tests.test", "true" ]}, then: 0, else: 1}}}},
  {$group: {_id: "$_id", failedCount: {$sum: "$failed"}, acts: {$push: "$acts.tests"}}}
])

但我无法让它反转 $unwind 操作，我只能得到与原始数据结构不同的结果数据结构。是否有可能使结果看起来与原始集合完全一样，但具有新的聚合值？

/gemigspam

Answer 1

有一个特殊的技巧来处理这个问题，但首先如果你有 MongoDB 2.6 或更高版本，那么你实际上可以做你想做的事而不需要使用 $unwind。 .如果您正在处理大量文档，这对于性能来说非常方便。

这里的关键运算符是$map它处理数组和 $allElementsTrue将评估您的“结果”字段的运算符。这里使用“map”既可以测试内部的“tests”数组，也可以查看其中的“result”字段是否都满足真实条件。在外部数组的情况下，这个“结果”可以根据需要放入那些文档中，当然文档的完整评估遵循相同的规则:

db.test.aggregate([
    { "$project": {
        "name": 1,
        "result": {
            "$allElementsTrue": {
                "$map": {
                    "input": "$acts",
                    "as": "act",
                    "in": {
                        "$allElementsTrue": {
                            "$map": {
                                 "input": "$$act.tests",
                                 "as": "test",
                                 "in": "$$test.result"
                            }
                        }
                    }
                }
            }
        },
        "acts": {
            "$map": {
                 "input": "$acts",
                 "as": "act",
                 "in": {
                    "name": "$$act.name",
                    "result": {
                        "$allElementsTrue": {
                            "$map": {
                                "input": "$$act.tests",
                                "as": "test",
                                "in": "$$test.result"
                            }
                        }
                    },
                    "tests": "$$act.tests"
                 }
            }
        }
    }}
])

在早期版本中执行此操作的方法需要您 $group分两步返回，以便在再次对这些“结果”字段进行测试时“重建”数组。这里的另一个区别也是使用 $min operator as false 将被视为比 true 更小的值，并计算出相同的“allElements”概念:

db.test.aggregate([
    { "$unwind": "$acts" },
    { "$unwind": "$acts.tests" },
    { "$group": {
        "_id": {
            "_id": "$_id",
            "name": "$name",
            "actName": "$acts.name"
        },
        "result": { "$min": "$acts.tests.result" },
        "tests": {
           "$push": {
               "name": "$acts.tests.name",
               "result": "$acts.tests.result"
           }
        }
    }},
    { "$group": {
        "_id": "$_id._id",
        "name": { "$first": "$_id.name" },
        "result": { "$min": "$result" },
        "acts": {
            "$push": {
                "name": "$_id.actName",
                "result": "$result",
                "tests": "$tests"
            }
        }
    }}
])